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eHC^ 


1 


WENTWORTH'S 
SERIES    OF     MATHEMATICS 


First  Steps  in  Number. 

Primary  Arithmetic. 

Grammar  School  Arithmetic 

High  School  Arithmetic. 

Exercises  in  Arithmetic. 

Shorter  Course  in  Algebra. 

Elements  of  Algebra.  Complete  Algebra. 

College  Algebra.  Exercises  in  Algebra. 

Plane  Geometry. 

Plane  and  Solid  Geometry. 

Exercises  in  Geometry. 

PI.  and  Sol.  Geometry  and  PI,  Trigonometry. 

Plane  Trigonometry  and  Tables. 

Plane  and  Spherical  Trigonometry, 

Surveying. 

PI.  and  Sph.  Trigonometry,  Surveying,  and  Tables 

Trigonometry,  Surveying,  and  Navigation. 

Trigonometry  Formulas. 

Logarithmic  and  Trigonometric  Tables  (Seven). 

Log.  and  Trig.  Tables  (Complete  Edition). 

Analytic  Geometry, 


Special  Terms  and  Circular  on  Application, 


PLANE  AND  SPHEEICAL 


TRIGONOMETRY, 


AND 


SURVEYING 


BY 


G.  A.  WENTWORTH,  A.M.,   ^ 

PBOFEBSOR  OF  JE^^MATICS  IN  PHILLIPS  EXETER  ACADEJttT. 


2Ceacl)ers*  lEtiitton* 

BOSTON  COLLEGE 
PHYSICS  DEPT. 

BOSTON,  U.S.A.: 
GINN  &  COMPANY,   PTTBLISHEES. 

1891. 


D/^ 


-y  Jj   t 


X 


Entered,  according  to  the  Act  of  Congress,  in  the  year  1886,  by 

G,  A.  WENTWORTH, 

in  the  Office  of  the  Librarian  of  Congress,  at  Washington. 


h4 


Typography  by  J.  S.  Gushing  &  Co.,  Boston,  U.S.A. 


Prksswork.  by  Ginn  &  Co.,  Boston,  U.S.A. 


JUN19 


/ 


PREFACE. 


THIS  edition  is  intended  for  teachers,  and  for  them  only.  The 
publishers  will  under  no  circumstances  sell  the  book  except  to 
teachers  of  Wentworth's  Trigonometry  ;  and  every  teacher  must  con- 
sider himself  in  honor  bound  not  to  leave  his  copy  where  pupils  can 
have  access  to  it,  and  not  to  sell  his  copy  except  to  the  publishers, 
Messrs.  Ginn  &  Company, 

It  is  hoped  that  young  teachers  will  derive  great  advantage  from 
studying  the  systematic  arrangement  of  the  work,  and  that  all  teach- 
ers who  are  pressed  for  time  will  find  great  relief  by  not  being 
obliged  to  work  out  every  problem  in  the  Trigonometry  and  Sur- 
veying. 

G.  A.  WENTWORTH. 


^\ 


TEIGONOMETET. 


Exercise  I.     Page  5. 


1.  What  are  the  functions  of  the 
other  acute  angle  B  of  the  triangle 
ABC{Yig.  2)? 


sin  B  =  -1 
c 

cos  5  = 

tan  B  =  -, 
a 

cot  5  = 

sec  B  =  -, 
a 

CSC  B  = 

2.  Prove  that  if  two  angles,  A 
and  B,  are  complements  of  each 
other  (i,e.,  ifA  +  B  =  90°),  then 

sin  A  =  cos  B,     cos  A  =  sin  B, 


tan  A  =  cot  B, 

sec  A  =  esc  B, 

Bin  A  =  -y 
c 

COS  A^-> 
c 

tan^==  -, 


cot  A=-, 
a 


cot  A  =  tan  B, 

CSC  A  =  sec  B. 


(1-) 


cos  B 


sec  A  = 


esc  J.  =  — 
a 


sin  B=~, 
c 

cot  B  =  -, 


tan  B  =  -, 

a 

CSC  iJ  =  -, 


sec  5  = 


3.  Find  the  values  of  the  func- 
tions of  A,  if  a,  b,  c  respectively 
have  the  following  values : 

(i.)  3,    4,    5.  (iv.)  9,       40,     41. 

(ii.)  5,  12,  13.    (v.)  3.9,    8,       8.9. 

(iii.)  8,  15,  17.  (vi.)  1.19,  1.20,  1.69. 


(lii. 


(v.) 


sin  A 
cos  A  ■■ 
tan^ 
cot^  = 
sec  A  -■ 
CSC  A  -- 

)  sin  ^ : 
cos  A  = 
tan.A  = 
cot  J.- 
sec  J.  = 
CSC  A  = 

sin  A  -- 
tan^  = 
sec  A  = 


3 

— ) 
5 

4 

5' 

3 

4 

4 

3' 

5 

4 

5^ 

3 


17 
15 

15' 

15 

'  8' 

17 
15' 

17 


.39 

"89' 
39 
80' 
89 
80' 


(ii 

)  sin^ 

cos  A 

tan^ 

cot -4 

sec  A 

CSC  4 

(iv 

.)  sin  A 

COS,  A 

tan^ 

cot  J. 

sec  A 

CSC  A- 

cos  A 

COt^: 

CSC  A  ■■ 


13 

=  1? 
13' 

__5^ 

^12 

_12 

"  5' 

^13 

12 

13 

"  5* 

9 

^41' 

40 

Ti' 

_^ 
'40' 

40 

^T 

41 
^40' 
.41 
"  9' 

80 

89' 

80 

39' 

89 

39" 


TRIGONOMETRY. 


,  .,  .  ,  119 
( VI.)  Sin  A  =  — 1 
^     ^  169 

tan  A  =  - — ) 
120 

A      169 

sec  A  = > 

120 


A      120 
cos  A  = . 

169 

,   ,      120 

cot  A  = . 

119 


CSC  A 


169 
119" 


4.  What  condition  must  be  ful- 
filled by  the  lengths  of  the  three 
lines  a,  b,  c  (Fig.  2)  in  order  to 
make  them  the  sides  of  a  right  tri- 
angle? Is  this  condition  fulfilled 
in  Example  3  ? 

a^  +  b^  =  c^. 

5.  Find  the  values  of  the  func- 
tions of  A,  if  a,  h,  c  respectively 
have  the  following  values  : 

(i.)  2mn,  m^  —  n'^,  mn? -\- n^. 


(ii.)  _^,    x-vy, 
x-y 

x'^  +  y'^ 
x-y 

(iii.)  pqr,  qrs,  rsp. 

/■    s  mn     mv     nr 

<iv.) 


•pq^       sq      ps 


(i.) 

.      a        2mn 
sm  A  =  -  = 


mr  -|-  n'' 


cos  .A  =  -  = 


mr 


i.       A      a        2mn 
tan  A  =  -  = ) 


m''  —  n" 


cotA  =  -  = 


m^ 


2mn 


e      Wr  +nr 


see  A  =  -=^ 

0      7\i?-  —vi^ 

A     c     m?  -f-  w? 

CSC  4  =^  =  *^-^-i ^• 

^  a       2mn 


(ii.) 


2x]i_ 


X  —  y     x^  +  y'^ 

x^  -f  y'^^ 

cos  A  = 

x^  4-  y^ 

__x^  —  y^ 
o;''^  -F  y^' 

tan  A- 

_  2a;.v  ..       1 
x—y       X  -v  y 

2xy 
x^  -  3/2' 

cot  J.= 

2a'2/ 

_x'^  —  y"^ 
2xy  ' 

sec  A-- 

1     ..  a;2  +  y2 
X  ■}■  y       x  —  y 

_x'^  +  y'^ 
x^  —  y'^ 

csc  A-- 

^  —  Vs,  x"^  +  y"^ 
2xy        x  —  y 

__'3?  -\-  y"^ 
2xy 

(iii.) 


pqr      q 

sm  A  = =  -' 

rsp      s 

pqr     p 

tan  A  =  — ■  =  -' 
qrs      s 

rsp      p 
sec  A  = =  -' 


qrs 


qrs       q 

cos  A=  — ■  =-■> 
rsp      p 

qrs      s 

cot  J.  = =  -' 

pqr     p 

rsp      s 
CSC  A  =  —  =  -• 
2W      q 


(iv.) 


sm  J.  =  —  X  =^-  =  — , 
pq       nr       qr 

mv       ps      mpv 
cos  A  =  —  X  ^—  =  ^-- 
nr 


sq 


nqr 


.        n      sq      ns 

tan  J.  =  —  X  -^  =  --. 

pq      mv     pv 

mn      sq      ns 


,       sq       nr 
sec  A  ==  -^  X  — 


nqr 
mv       ps      mpv 


,€se 


Ti^n  ^  ps     ms 


TEACHERS     EDITION. 


6.  Prove  that  the  values  of  a,  h, 
c,  in  (i.)  and  (ii.),  Example  5,  satisfy 
the  condition  necessary  to  make 
them  the  sides  of  a  right  triangle. 

(i-) 
a2  _^  52  ^  g2^ 

(2  mnf  +  (m^  —  n^Y  =  (m'^  +  n^)^, 

4  rn^n^  +  m*  —  2  m?v?  +  n* 

=  m*  +  2  m^v?  +  n*, 


2xy\ 


(li.) 


3^     +(aj  +  2/)'  = 


x-yj 


x^  +  3/^ 


•2      o  -  +  x2  +  2a;2/+2/2 

x^  —  Axy  ■\-  y^ 

_a;*+2a;V+.V^ 
a;''  —  2  a;?/  +  2/2 
4a;y +  a;*-2a;y +  2/* 

=  a;*  +  2cey +  2/*, 
a*  +  2a;y  +  2/*  =  a;*  +  2a;y  +  2/*. 

7.  What  equations  of  condition 
must  be  satisfied  by  the  values  of 
a,  5,  c,  in  (iii.)  and  (iv.),  Example  5, 
in  order  that  the  values  may  repre- 
sent the  sides  of  a  right  triangle. 

(iii.) 

^2g2^2  _|_  g2j,2g2  _  7'2g2p2^ 

or     Jp^<f'  +  2^8^  =pH'^. 


(iv.) 


t2/>-)2 


2/1 12 


1 


or     m^n's^  +  m^p^v^  =  n^(fr^. 

8.   Compute  the  fun<;tions  of  J. 
and  B  when  a  =  24,  &  =  143. 


c  =  \/(24)2  +  (143)2 


=  \/21025 
=  145. 


24 
sin  J.  =  —  =  cos  B, 

145 

J      143       .     „ 
cos  A  =  —  =  sm  B, 
145 

94 

tan  J.  =  —  =  cot  B, 

143 

cot  J.  =  —  =  tan  B, 
24 

A     145  o 

sec  A  =  — •  =  CSC  B, 
143 

A         145  r> 

CSC  -4  =  —  =  sec  B. 

24 


9.  Compute  the  functions  of  A 
and  B  when  a  =  0.264,  c  =  0.265. 

^2  =  g2  _  0^2 

=  0.070225-0.069696 
=  0.000529. 
.-.  6  =  0.023. 

sin^  =  ^  =  ?^  =  cos5, 
c      265 

cos  ^  =  -  =  —  =  sin  B, 
c      265 

tanJl  =  ?  =  ^  =  cot5, 
6       23 

cot  .4  =  -  =  —  =tan5, 
a     264 

sec^  =  ^  =  ^  =  csc5, 
6       23 

csc^  =  ^  =  ?^  =  sec.g. 
a     264 

10.  Compute  the  functions  oi  A 
and  B  when  6  =  9.5,  c  =  19.3. 

a^  =  c^  —  52 

=  372.49  -  90.25 

=  282.24. 
.-.  a  =  16.8. 

sin.4=^=l^  =  cos5, 
c      193 


TRIGONOMETEY. 


.      5       95         .J, 
cos  A=  -  =  —  =  sm  ij, 
c      193 

tan  A  =  -  =  — ■  =  cot  B, 
b       95 


cot  ^  =  - 
a 


95 
168 


tan  5, 


sec  ^  =  -  =  —  =  CSC  is, 
b      95 

.     c      193  D 

CSC  ^  =  -  =  —  =  sec  B. 
a     168 


11.   Compute  the  functions  of  A 
and  B  when 


a  =  Vp^  +  g^,  b  =  V2pq 

p^  +  2pg'  +  ^-^  =  c'^. 
p  +  q  =  c. 
a 


sin  A  = 


cos  A  =  -  = 

c     p  +  q 


p  +  q 

\/2pq 


cos  5, 


sin  B, 


0 


V2pq 
^2pq 


cot  J.  =  -  =      

^      y/p"^  +  q^ 

o      y/2pq 


=  tan  B, 
=  esc  B, 


CSC  ^ 


c  _    p  -\-  q     _ 


^      y/p^  +  (^ 


=  sec  5, 


12.   Compute  the  functions  of  A 
and  B  when 


a=Vp'^  +pq,  c=p  +  q, 
b^  =  c^  —  Q? 
=  (f  -{-pq. 
.'.  b  =  Vq^  +  pq. 

■      A      ^      Vp2  +pq  ^ 

sm  A  =  -=  —^ i-2  =  cos  B, 

c       P  +  q 


cos 


.      b      y/f  +pq        . 

A  =  -  =  — i i-i  =  sm  B, 

c         P  +  2 

"      Vq^  +  pq 


«      \/pi  -j-pq 

A  ^  P  +  Q  T> 

sec  A  =  j-  =    f      ^  -  =  CSC  B, 
"      Vq^  +  pq 


CSC  A  = 


c  _     p  +  q 


=  sec  B. 


^      Vp^  +  pq 

13.   Compute  the  functions  of  A 
and  B  when 

b  =  2->/pq,  c  =p  +  q. 
a2  +  52  =  c\ 
a?  +  ipq  =p'^  +  2pq  +  q^, 

a  =p  —  q' 

sm  tI  =  -  =  ^ — ^  =  cos  B, 

c      p  +  q 

.      b     2Vpq       .     „ 
cos  J.  =  -  =  — i-i  =  sm  B, 
c      p  +  q 


tan -4 


cot  J.  =  -  = 
a     p 


pj-q 
^y/pq 
2V^q 


sec  J.  ==  ^ 
0 


CSC  A  =  - 
a 


9. 

P  +  9. 


pq 

.P  +  9 
p-q 


cot  B, 

tan  5, 
CSC  B, 

=  sec  B. 


( 


14.   Compute  the  functions  of  A 
when  a  =  2  6. 

a  =  2&, 
a2  +  62  =  c2, 
462  +  62_c2, 
562  =  c2, 

C-6V5. 


TEACHERS     EDITION. 


sm  -d.  =  -  = 
CO?  A  = 


a     _2b_ 


=  f  V5  =  0.89443, 


bV5 


2. 


b        b 

cotA  =  -  =  h 
a     2 

b        b 

cscA  =  '-  =  ^'=iV5. 
a       2b       ^ 


15.   Compute  the  functions  of  J. 


when  a  =  I  c. 


a  =  fc, 

52  ^  ^2  _  0^2^ 

6  =  Vc^  —  a^ 


16.   Compute  the  functions  of  A 
when  a  +  b  =  ^c. 

a2  +  J2  _  g2^ 

a2  +  62  +  2a5  =  ffc2, 
a2-2a6  +  62  =  _7_c2, 


2 


sm  ^  =  -  =  — 
p      3 
0      2 


-  =  -, 
a     3' 


&      2^ 
cosul  =  -  =  -T —  =  ^\/5, 


tan  J.  =  -  = 


^     ^V5 


cot^  =  ^  =  :^. 


fa 


=  fV5, 


sec  ^  =  - 

2 

csc^  =  ^  =  ?. 
a     2 


^"'   =fV5, 


a  —  b  = 

1^- 

a  +  b  = 

=i«. 

26  = 

=  fo-|V7, 

6  = 

=  1-1  V7, 

2a  = 

4 

a  = 

=  fc+|V7, 

c 

a 

8     5+ V? 

8a 


c  = 


sm  ^  =  -  = 
c 


a 


5+ V7 
5  + V7 


8a 


6  + V7 
c 


6._£  V7 


cos  ^  =  -  = 
c 


,        .     a     5+v^ 

tan  J.  =  T  = ^' 

^     6 -a/7 

cot^  =  5  =  5-^ 


5-V7 


a     5+^7 


sec  -4  =  -  = 


b     5-V7 


esc  ^  =  -  = 


a     5+y^ 


.'.^ 


6 


TRIGONOMETRY. 


17.   Compute  the  functions  of  J. 
when 


a  —  6  =  — ' 
4 


a2-2a5  +  Z.2 


16 

+  52  _   c2 


2ab 


15c^ 
16 


+  &2 


a2  +  2a6  +  62  = 


a  +  6  =  -  Vsl, 
4 


31  c^ 
16 


I.      I' 
a  —  0  =  -, 

4 

2a  =  -V31+-- 
4  4 

..  a  =  -(\/31  +  l). 

2  6  =  -  VSI  -  -. 
4  4 

.-.  6  =  |(V3l-l). 
^(V3l  +  1) 


CSC  ^  =  -  = 


«    V3I  +  1 


18.   Find    a    if    sin  J.  =  |    and 


G  =  20.5. 


sin^  =  ^  =  ^. 
c      5 

g    ^3 

20.5     5' 

5a  =  61.5, 

a  =12.3. 


19. 

Find 

h  if  cos 

A  = 

=  0.44 

and 

c  =  3.5 

h 
c 

h 

3.5 

.-.  h 

=  0.44, 

=  0.44. 
=  1.54. 

Sin  J.  =  -  = 
c 


h     8 


^(V31-l) 


cos  A  =  —  = 
c 


\/3l  +  l 


V31-1 


tanyJ=g_V31+l, 
^      V31-1 

,    ,      h      V31-1 
cot  A  =  -  = 1 

«     Vsl  +  i 


sec  A  =  -  = 


V31-1 


20.  Find   a   if    tan  J.  =  J^  and 

/>  —  2-5- 

'^"  11 

a  _  _a_  _  11 

6~"2^~  a" 
.    lla_ll 
"27        3' 

a  =  9. 

21.  Find    5    if   cot  J.  =  4    and 

a  =  17. 

5  =  A  =  4 
a     17 

.-.  &  =  68. 

22.  Find    c    if   sec  J.  ==  2    and 
&  =  20. 

^  =  —  =2 

6  20        ■ 

.-.  c  =  40.  . 


TEACHERS     EDITIOISr. 


6.45  and 


23.   Find   c  if  esc  J. 
a  =  35.6 

.    CSC  J.  =  -  =  -^  =  6.45. 
a     35.6 

.-.  c  =  229.62. 


24.  Construct  a  right  triangle ; 
given  c  =  Q,  tan  -4  =  f . 

tan^  =  -. 

.*.     a  =  3  and  &  =  2. 

Draw  AB  -=  2,  and  .5C  ±  to  ^5 
=  3  ;  join  Cand  A. 

Prolong  vie  to  D,  making  AD  =  6. 

Draw  DE 1.  to  AB  produced. 

Rt.  A  ADE  will  be  similar  to  rt. 
^ACB. 

.'.  ADE  is  the  rt.  A  required. 


25.  Construct  a  right  triangle  ; 
given  a  =  3.5,  cos  A  =  ^. 

Construct  A  A^B'C  so  that  y=l, 
c^=  2.     Then  cos  A  =  >}. 

Construct  A  ABC  similar  to 
A^B'C^,  and  having  a  =  3.5. 

26.  Construct  a  right  triangle ; 
given  5  =  2,  sin  J.  =  0.6. 

Construct  rt.  A  A^B^C^,  m'aking 
a^=  6,  and  c  =  10. 

Then  sin  A^=  j%. 

Construct  A  ABC  similar  to 
A^B'C^,  and  having  6  =  2. 

27.  Construct  a  right  triangle; 
given  &  =  4,  cscA  =  4. 

Construct  rt.  A  A^B^C'',  having 
c''=  4  and  a^=  1. 

Then  construct  A  J.5(7  similar  to 
A  A^B'C,  and  having  6  =  4. 


28.  In  a  right  triangle,  c  =  2.5 
miles,  sin  J.  =  0.6,  cos  A  =  0.8;  com- 
pute the  legs. 


•          A          ^ 

sm  A=  — 
e 

A       ^ 

cos  A-=  — 
c 

a  =  c  sin  A. 

.'.  b  =  c  cos  A 

a  =1.5. 

.-.  6  =  2. 

30.  Find,  by  means  of  the  table, 
the  legs  of  a  right  triangle  ifA=  20°, 
c  =  1 ;  also,  if  ^  =  20°,  c  =  4. 

A  =  20°,  c  =  1. 


sm  A  =-. 
c 


cos  ^ 


.•.  a  =  c  sin  ^.  .*.  6  =  c  cos  J.. 

.-.  a  =  0.342.  .-.  6  =  0.940. 

^  =  20°,  c  =  4. 

.-.  a  =  4x0.342  .-.  6  =  4x0.940 
=  1.368.  =3.760. 

31.  In  a  right  triangle,  given 
a  =  3  and  c  =  5  ;  find  the  h3^pote- 
nuse  of  a  similar  triangle  in  which 
a  =  240,000  miles. 

a:  C::  240,000  :  x, 
3:5::  240,000  :  x. 
.-.  a;  =  400,000. 

32.  By  dividing  the  length  of  a 
vertical  rod  by  the  length  of  its 
horizontal  shadow,  the  tangent  of 
the  angle  of  elevation  of  the  sun  at 
the  time  of  observation  was  found 
to  be  0.82.  How  high  is  a  tower, 
if  the  length  of  its  horizontal  shadow 
at  the  same  time  is  174.3  yards  ? 

tan^  =  -  =  0.82. 
6 

.-.  a  =  0.82  6. 

b  =  174.3  yards. 

.-.  «  =  0.82  of  174.3  yards 

=  142.926. 


TRIGONOMETRY. 


Exercise  II.     Page  8. 


1.  Eepresent  by  lines  the  func- 
tions of  a  larger  angle  than  that 
shown  in  Fig.  3. 


Fig.  1. 

2.  Show  that  sinx  is  less  than 
tana;. 

In  Fig.  1,  OM:  PM::OA:  AS, 
but  0M<  OA. 

.'.  PM<  AS. 

3.  Show  that  sec  x  is  greater  than 
tana;. 

0/S'=sec,     AS=id.n. 

In  rt.  A  OAS,  Hyp.  OaS'>  side  AS 

.'.  sec  >  tan. 

4.  Show  that  esc  x  is  greater  than 
cot>r. 

OT=csc,     BT=cot 

In  A  BOT,  Hyp.  0T>  side  BT. 

.'.  CSC  >  cot. 

5.  Construct  the  angle  x  if  tan  x 

Let  O  BAM  he  a  unit  circle,  with 
centre  0;  then  construct  AT  tan- 
gent to  the  circle  at  J.  =  3  OA  ; 
then  J. or  is  required  angle. 


Construct  the  angle  x  if  esc  x 


Let  O  ABM  be  a  unit  circle,  with 
centre  0;  construct -ST tangent  to 
the  circle  at  B  =  2  OA  ;  connect 
OT;  then  J. OT is  required  angle. 

7.  Construct  the  angle  a;  if  cos  x 

—  2^- 

Take  0M=  I  radius  OA.  At  M 
erect  a  ±  to  meet  the  circumference 
at  P.     Draw  OP. 

Then  is  POif  the  angle  required. 

8.  Construct  the  angle  a;  if  sin  x 
=  cos  X. 

Let  PM=  sin  x  and  0M=  cos  x. 
But,  by  hypothesis,  PM--=  OM. 
.•.  by  Geometry,  x  =  45°. 
Hence,  construct  an  Z  45°. 

9.  Construct  the  angle  a;  if  sin  x 
=  2  cos  X. 

Construct  rt.  Z  PMO,  making 
PM=  2  OM.     Draw  OP. 

Then  POM\b  the  angle  required. 

if 


10.  Construct    the    angle 
4  sin  X  =  tan  x. 

Take  J  of  radius  OA  to  M.  At 
J/  erect  a  _L  to  meet  the  circumfer- 
ence at  P.     Draw  OP. 

Then  POil/ is  the  required  angle. 

11.  Show  that  the  sine  of  an 
angle  is  equal  to  one-half  the  chord 
of  twice  the  angle. 

Have  given  Z  POA. 

Construct  POB  =  2  POA.  Draw 
chord  PB.  Then  it  is  ±  to  OA  • 
and  PM,  its  half,  is  the  sine  of  POA. 

.•.  sin  a;  =  2  chord  2a;. 


TEACHEES     EDITION. 


9 


12.  Find  x  if  since  is  equal  to 
one-half  the  side  of  a  regular  in- 
scribed decagon. 

Let  AC\)Q  a  side  of  a  decagon. 
360° 


Then 


10 


=  36°  or^Oa 


Draw  OB  bisecting  AO.  Then 
Z ^0(7  will  be  bisected,  and  Z  ^05 
=  18°. 

But  the  sine  oiAOB=^\AQ. 

.'.  X  or  AOB  =  18°. 

13.  Given  x  and  y  {x  +  y  being 
less  than  90°) ;  construct  the  value 
of  sin  {x  +  y)  —  sin  x. 

Let  AB  =  sin  (a;  +  y)  in  a  circle 
whose  centre  is  0,  and  CD  =  sin  x. 

Then,  with  a  radius  equal  to  CI), 
describe  an  arc  from  B,  as  centre, 
cutting  AB  &t  E. 

Then  BJA  will  be  the  constructed 
value  of  sin  (x  +  y)  —  sin  x. 

14.  Given  x  and  y  (x  +  y  being 
less  than  90°) ;  construct  the  value 
of  tan  (x  +  y)  —  sin  {x  +  y)  +  tan  x 
—  sin  X. 

Let      AB  =  sin  {x  +  y), 
and  CI)  =  sin  a; ; 

also  EF  =  tan  {x  +  3/), 

and  0F=i2,xxx. 

From  F  with  a  radius  =  AB  take 

From  ^with  a  radius  =  GF  add 
HI. 

From  /  with  a  radius  =  CD  take 

Then  ^^will  be  the  constructed 
value  of  tan  {x  +  y)  —  sin  {x  +  y)  + 
tan  X  —  sin  x. 

15.  Given  an  angle  x  ;  construct 
an  angle  y  such  that  siny  =  2  sin  a?. 


Let  AB  be  the  sine  of  the  Z  a;  in 
a  circle  whose  centre  is  0. 

Draw  AC  perpendicular  to  the 
vertical  diameter. 

Then  CO  =  AB. 

Take  CF  on  vertical  diameter 
=  CO.  Draw  FD  perpendicular  to 
vertical  diameter,  and  meeting  cir- 
cumference at  D. 

Draw  DE  perpendicular  to  OB 
and  draw  OD. 

0F=  2  CO  by  construction. 

ED=FO;  FO  being  the  projec- 
tion of  the  radius  OD. 

.-.  DE  =  2  AB,  and  DOB  =  &nglQ 
required. 

16.  Given  an  angle  x  ;  construct 
an  angle  y  such  that  cos  3/  =  J  cos  a;. 

Let  OB  =  cos  AOB. 

Erect  a  ±  CD  at  C,  the  middle 
point  of  OB,  and  meeting  the  cir- 
cumference at  D.     Draw  DO. 

Then  DOB  is  the  angle  required. 

17.  Given  an  angle  x  ;  construct 
an  angle  y  such  that  tan  3/ =  3  tan  a;. 

Let  AB  be  the  tangent  of  x. 

Prolong  AB  to  C,  making  AC= 
3  AB,  and  draw  OC  from  0,  the 
centre  of  the  circle. 

COA  is  the  required  angle. 

18.  Given  an  angle  x  ;  construct 
an  angle  y  such  that  sec  y  =  esc  a;. 

Since  sec  =  esc, 

£  —  £ 
b      a 

.'.  a  =  b. 

Hence,  construct  an  isosceles  right 
triangle. 

The  required  angle  will  be  45°. 


10 


TRIGONOMETRY. 


19.  Show  by  construction  that 
2  sin  J.  >  sin  2  A. 

Construct  Z  BOC  and  Z  COA 
each  equal  to  the  given  Z  A. 

Then  AB  =  2  sin  A,  and  AD,  the 
J_  let  fall  from  A  to  OB,  =  sin  2  ^. 
But  AB  >  AD. 

Hence  2  sin  J.  >  sin  2  A. 

20.  Given  two  angles  A  and  B 
{A^  B  being  less  than  90°),  show 
that  sin  {A  +  B)  <  sin  A  +  sin  B. 

Construct  HOK  =  Z  ^,  and  COH 
=  ZB. 

Then  sin  (A  +  B)  =  CP,  sin  A  = 
UK,  sin  ^  =  CD. 

Now  CP<CD  +  DE, 

and  EK>DE. 

.-.  (7P<  CD+HK. 


.-.  sin  ( J.  +  P)  <  sin  ^  +  sin  B. 

21.   Given  sin  a;  in  a  unit  circle ; 
find  the  length  of  a  line  corre^^^'^  ^ 
ing  in  position  to  sin  a;  11       .„uio 
whose  radius  is  r. 

1  :  r  :  :  sin  x  :  required  line. 

.•.  length  of  line  required  =  r  sin  x. 


22.  In  a  right  triangle,  given  the 
hypotenuse  c,  and  also  sin  A  =  7?i, 
cos  A  =  n  ;  find  the  legs. 

sm  ^  =  -  =  m. 
c 

.'.  a  =  cm. 


cos  A  —  -  =  n. 
c 

.'.  b  =  en. 


Exercise  III.     Page  11. 


1.  Express  the  following  func- 
tions as  functions  of  the  comple- 
mentary angle : 

sin  30°.  CSC  18°  10^ 

cos  45°.  cos37°24^ 

tan  89°.  cot82°  19^ 

cot  15°.  CSC  54°  46^ 

sin  30°  =  cos  (90°  -  30°)  =  cos  60°. 
cos  45°  =  sin  (90° -45°)  =  sin  45°. 
tan  89°  =  cot  (90° -89°)  =  cot  1°. 
cot  15°  =  tan  (90° -15°)  =  tan  75°. 
esc  18°  10^=  sec  (90°  -  18°  10^ 
=  sec  71°  50^. 

cos  37°  24^=  sin  (90°  -  37°  24^ 

=  sin  52°  36^. 
cot  82°  19^=  tan  (90°  -  82°  19^ 

=  tan  7°  41^ 
esc  54°  46^=  sec  (90°  -  54°  46^ 

=  sec  35°  14^ 


2.  Express  the  following  func- 
tions as  functions  of  an  angle  less 
than  45° : 


sin  60°. 

CSC  69°  2^. 

cos  75°. 

cos  85°  39^ 

tan  57°. 

cot  89°  59^ 

cot  84°. 

CSC  45°  V. 

sin  60°  =  cos  (90°  -  60°)  =  cos  30°. 
cos  75°  =  sin  (90°  -  75°)  =  sin  15°. 
tan  57°  =  cot  (90°  -  57°)  =  cot  33°. 
cot  84°  =  tan  (90°  -  84°)  =  tan  6°, 
esc  69°  2^  =  sec  (90°  -  69°  2') 

=  sec  20°  58^. 
cos  85°  39^=  sin  (90°  -  85°  39^ 

=  sin  4°  21^. 
cot  89°  59^=  tan  (90°  -  89°  590 

=  tanO°l^ 
esc  45°  V  =  sec  (90°  -  45°  1^ 

=  sec  44°  59^ 


TEACHEES     EDITION. 


11 


3.  Given    tan  30°  =  ^  V3  ;    find 
cot  60°. 

tan30°  =  cot(90°-30°) 
=  cot  60°. 
.•.cot60°  =  J\/3, 

4.  Given  tan  A  =  cot  A  ;  find  A. 

tan  A  =  cot  (90°  -  A), 
90°  -  J.  =  A, 
2^  =  90°. 
.-.  A  =  45°. 

5.  Given  cos  ^  =  sin  2  J. ;  find  J.. 

cos  A  =  sin  (90°  -  A), 
dO°-A  =  2A, 
3^  =  90°. 
.-.  A  =  30°. 

6.  Given  sin  ^  =  cos  2 .4 ;  find  A. 

sin  A  =  cos  (90°  -  A\ 
90°-A  =  2A, 
3  A  =  90°. 
/.  A  =  30°. 


7.    Given  cos  J.  =  sin  (-15°  —  J  ^  ; 
find  J.. 

cos  A  =  sin  (90°  -  A), 
90°- J.  =  45° -J  J., 
180° -2^  =  90° -A 
.-.  A  =  90°. 


8.  Given  cot  ^  J.  =  tan  A ;  find  A. 

tan  A  =  cot  (90°  -  A), 
J^  =  90°-^, 

^  =  180° -2^, 
3^  =  180°. 
.-.  A  =  60°. 

9.  Given  tan  (45°  +  ^)  =  cot  ^ ; 
find  A 

cot  A  =  tan  (90°  -  A), 
tan  (90°  -A)  =  tan  (45°  +  A), 
90°  -  J.  =  45°  +  A, 
2  J.  =  45°. 
.'.  A  =  22°  30^ 

10.  Find  A  if  sin  ^  =  cos  4  A. 

sin  A  =  cos  (90°  -  A), 
dO°-A  =  4:A, 
5  A  =  90°. 
.'.  A  =  18°. 

11.  Find  ^  if  cot  ^  =  tan  8^. 

cot  A  =  tan  (90°  -  A), 
8A  =  90°-A, 
9  A  =  90°. 

.'.  A  =  10°. 

12.  Find  ^  if  cot  J.  =  tan  nA. 

cot  A  =  tan  (90°  -  A), 
90°- A  =  nA, 

90°  =  A{n+  1). 

.:A  =  ^^. 
n  +  1 


Exercise  IV.     Page  12. 


1.  Prove  Formulas  [1]  -  [3],  using 
for  the  functions  the  line  values  in 
unit  circle  given  in  §  3, 


[1].  sin^^  +  cos^yl  =  1. 

[2].  tan^=^-HL4. 
cos  J. 


12 


TEIGONOMETRY. 


[3].  sin  AxcscA  =  l, 
cos  AxsQc  A  =  l, 
tan  J.X  cot  J.  =  l. 


Fig.  2. 

[1].  DC=  sin  J., 

BD  =  cos  J., 
DC^  +  BD^  =  (7^2 . 
but  CB^  =  1. 

.-.  BC^  +  BD""  =  1. 
.'.  sin^A  +  cosM  =  1. 


[2] 


BC=  sin  A, 
BB  =  cos  A, 
EF=iQ.nA. 
A  FBE  and  ^Ci)  are  similar. 
.-.  FE :  BE::  CB  :  BB. 
FE_CB_. 
BE    BB' 
but  BE==1. 

CB 
BB 
sin  A 


Or. 


FE  = 


tan  J.= 


cos  A 


[3].  CB  =  sin  J., 

55"=  CSC  A. 
In  similar  /i^  HGB  and  (75i), 
^S" :  QB  ::  BC:  CB. 


.    BH^BC ^ 
■  *  (?5      CX) ' 

but  (?5  =  1, 

BC^l. 

BH=—. 
CB 

BHxCB  =  l, 

CSC  J.  X  sin  J.  =  1. 

cos  A  =  BB, 
sec  J.  =  BF. 
In  similar  A  5i^^  and  BCB, 
BF:  BE::  BC :  BB. 
BF  _BC  . 
BE     BB' 
but  BE=1, 

BC^l. 

,'.  BF=^  ~ 
BB 

BFxBB  =  l, 

sec  A  X  cos  J.  =  1. 

tan  A  =  EF, 
cotA=GR. 
In  similar  A  GSB  and  FEB, 
OH:  OB::  BE:  FE, 

GH_BE. 
GB      FE' 
but  GB  =  1. 

BE==1. 
1 

FE 

GBxFE=l. 

cot  A  X  tan  A  =  l. 

2.   Prove  that  1  +tan2J.  =  sec'A 

tan^  =  7,  sec^  =  -. 

b  c 

a-'  +  b^  =  c\ 
Dividing  all  the  terms  by  5^, 


GH- 


TEACHERS     EDITION. 


13 


a^      52  ^  g2 

Substituting  for  —  and  —  their 

values  idiU^A  and  sec^J.,  we  have 
tan^^  +  1  =  sec^J.. 


3.    Prove  that  1  +  cot^^  -  csc^^. 

cot^  =-', 
a 

CSC  A  =  — 
a 

0,2  _j.  52  ^  c^. 
Dividing  all  the  terms  by  o?, 


O^        52  ^  g2 

a?      a?      0? 


Substitutins  for  —  and  —  their 
^  o?  a? 

values  cot^J.  and  csc^J.,  we  have 
1  +  cot2^  =  csc^J.. 


4.   Prove  that  cot  A  = 


cot^ 


cos  A 
sin  A 


sin  A  = 

a 

—  ~i 
c 

cos  J.  = 

h 
c 

Substituting,     _ . 

h 

a 

a 

c 

c 

.'.  cot  J.= 

cos 

A 

sin  A 


ExEECisE  V.    Page  14. 


1.    Find  the  values  of  the  other 
functions  when  sin  A  =  -^f. 

sin^J.  +  cos^J.  =  1, 

m2 

13 


cosM  =  1 

cos  A 


=v-(iy 


25 
169' 


•.  cos  J.  =  — • 
13 

tan^=^ 
cos 


12 
5" 


cot  A  is  reciprocal  of  tan  A. 

.'.  cot^  =  A. 
12 

sec  A  is  reciprocal  of  cos  A. 


.-.  sec  A  =  ^. 


CSC  A  is  reciprocal  of  sin  A. 
13 
12* 


.-.  CSC  A  =  ^^ 


2.   Find  the  values  of  the  other 
functions  when  sin  A  =  0.8. 

sin^J.  +  cos^J.  =  1, 

cos2^  =  1  -  (0.8)2, 

cos  A  =  Vl-0.64. 

.-.  cos  A  =  0.6. 

tan^  =  ^-iB  =  0^. 
cos      0.6 

.-.  tan^=  1.3333. 

0.6 


cot^  = 


0.8 


14 


TRIGONOMETKY. 


cot  ^  =  0.75. 

sec  A  =  — 
0.6 

sec  A  =  1.6667. 
J_ 
0.8' 

CSC  A  =  1.25, 


CSC  A  = 


3.   Find  the  values  of  the  other 
functions  when  cos  A  =  f  f . 

sin^  +  cos^  =  1, 


sm 


1 

3600 

J  121 
^'3721 

3721 

sin 

11 

cos 

60 

cot 

1 
tan 

_60 
ll" 

sec 

1 

cos 

_61 
60' 

1 

61 

5.    Find  the  values  of  the  other 
functions  when  tan  A  =  ^. 


GSC  =  — -  =  --• 


sm 


11 


4.    Find  the  values  of  the  other 
functions  when  cos  ^=  0.28. 

sin^  +  cos^  =  1. 

sin  =  Vl  -  (0.28)2  _  VO.9216. 

=  0.96. 
sin      0.96 


tan^ 


cos 


0.28 
1 


=  3.4285. 


tan     3.4285 


=  0.29167. 


sec 


1-  =  —  =  3.5714. 

0.28 


cos 

1 


esc  =-^=-'-=  1.04167. 


sin 


0.96 


tan  J.= 

^3 

.  cot  J.  = 

3 
^4 

tan  J.= 

_  sin  A 
cos  A 

4 

sin 

A 

3 

Wl- 

sin 

^A 

3  sin  ^  =  4VT~  smM, 
9  sin^A  =  16-16  sin^^, 
25  sin2^  =  16, 
5  sin  J.  =  4. 


sin  A  =  — 
5 


cos  A  = 
sec  A  = 
CSC  A  = 


sin  A 
tan^ 

1 

cos  A 

1 

sin  A 


6.   Find  the  values  of  the  other 
functions  when  cot  J.  =  1. 
cot  J.  =  1. 
.'.  tan  J.  =  l. 

sin  A 


tan -4  = 
1  = 


cos  A 
sin  A 

Vl  -  sin'' A 


sin  A  =  Vl  —  sinM, 


sm 


'^  =  1  -  sin2^, 


2  sin2^  =  1, 


sm 


'■A=^^. 


TEACHERS     EDITION. 


15 


COS  J.= 


tan  J. 


sec  A^ 


1 


CSC  ^  = 


COS  A     I V2 
1 


sin  A     \y/2 


=  V2. 


=  V2. 


7.   Find  the  values  of  the  other 
functions  when  cot  J.  =  0.5. 

0.5 


tan -4  = 
tan^  = 


cot  A 
sin  A 


2. 


cos -4 
2  cos  A  =  sin  A. 

4  cos^  J.  —  sin^^  =  0     (squaring) 
cosM  +  sin^^  =  1 

5  cos^  J.  =  1 
cos  J.  =  a/-  =  0.45. 


4cos2J.4-  4sin2J.  =  4 

4cos2^—    sin2^  =  0 

5sin2J.  =  4 

sin  A  =  -^-  =  0.90 
sec  A  = 
CSC  A  = 


'5 

1 

cos  A 

1 


sin  A 


=  2.22. 
=  1.11. 


8.   Find  the  values  of  the  other 
functions  when  sec  -4  =  2. 

cos  A  = =  -, 

sec  A     2 

sin  A  =  Vl  —  cos^^ 


-^l^h4 


.-.  sin  A  =  iVS. 

tan  A  = .  =  ^——  =  Vs. 

cos  A        1 

2 

1  1 


cot^  = 


CSC 


tan^      ^ 
sin-4 


iV3. 


9.   Find  the  values  of  the  other 
functions  when  esc  A  =  \/2. 

sin  J.=  —  =  1V2, 
V2 


cos  ^  =  Vl  -  (^  V2)2  =  vT^ 

=  v|=jA 

tan^=i^  =  l, 
^V2 

cot  ^  =  1  =  1, 
1       ' 

sec  A  =  r-^  =  V2. 


iV2 


10.   Find  the  values  of  the  other 
functions  when  sin  A  =  7n. 


cos  A  =  Vl  —  sinM  =  Vl  —  m^, 
sin  ^  m 


tan -4  = 


cot^ 


cos  A        Vl  —  TO^ 

w  Vl  —  m'^ 
\  —  w? 

1  1  —7n? 


tan  yl      mVl  — m'^ 


16 


TRIGONOMETRY. 


sec  A  = 
CSC  A  = 


1 


1 


cos  A      ^i  _  ^2 

1    ^  1 

sin  A     m 


11.   Find  the  values  of  the  other 
,■  ^         ■      A         2m 


Oill  yi  — ■ 

1  +m2 

cos  A  ==  Vl  - 

-  sin^. 

.'.  cos  A  =  a/i  - 

4m2 
l  +  2m2  +  m* 

=Ji^ 

-2m2  +  m* 

^1  +  2m'^  +  m* 
1  —  m^ 


l  +  m2 

taii^  = 

sin 

cos 

2m 
l-m^ 

cot  J.= 

1 

tan 

l-m2 
2m 

sec  A  = 

1 

1+m^ 

cos 

l-m^ 

rsr,  y1  = 

1 

l+m2 

sin 


2m 


12.  Find  the  values  of  the  other 
2'mn 


functions  when  cos  A  = 


m^  +  n^ 


sin 


A=Vl^ 


COS'' 


4  m^n^ 


m*  +  2  m^w^  +  ?i* 


m^  —  2m''n'-^  +  n* 
m*  +  2  m^Ti^  +  n* 


m^  —  n' 


sm 


m'' 


,,1  ^  10  2 

tan  A  =  ^^^^  =    ^ 

cos         2  mil 


,    .        1          2m?i 
cot  A  =  - —  =  —^ ; 


tan 

.        1        m?  ^-  V? 

sec  A  = •=  —^ 

cos          2  mn 


esc  A  = 


sm 


m^  +  n^ 
Tin?  —  n^ 


13.   Given  tan  45°  =  1 ;  find  the 
other  functions  of  45°. 


(1) 


sin  45° 
cos  45° 

sin  45° 


tan  45°. 
=  1. 


cos  45° 
(2)       sin'^  +  cos^  =  1. 
By  (1),  sin  45°  =  cos  45°. 
By  (2),  cos2  45°  +  cos2  45°  =  1. 
2cos2  45°  =  l, 

cos2  45°=-, 
2 

cos  45°  =  -ytt  =  |\/2. 


sin  45° 

=  JV2. 

cot  45° 

-       ^       -^   -1 

tan  45°     1 

sec  45° 

-^1     -V2. 
iV2 

CSC  45° 

iV2 

14.   Given  sin  30°  =  | ;    find  the 
other  functions  of  30°. 

sin'^  +  cos^  =  1.- 


TEACHERS     EDITION. 


17 


cos  30° 

=V-i 

=  A?  =  iV3. 

tan  30° 

cot  30° 

=  1  .V3. 
iV3 

sec  30° 

=  J^  =  JV3. 

^V3 

CSC  30° 

=  1  =  2. 

J 

15.    Given   csc  60°  =  f  V3 
the  other  functions  of  60°. 


find 


sm  =  ■ — , 

CSC 


)°  =  -^=-iV3. 


sin  60^  = ==  -2 

fV3 


cos  60°  =  Vl  -  sin2, 
COS  60°  =  Vl  -  (i  V3)2 


\  A        9. 


4      2 


JrVS 


tan  60°  =  ^^  =  Vs. 


cot  60°  =  —  =  I  Vs. 
V3 

sec  60°  =  i  =  2. 


16.   Given  tan  15°  =  2 -VS 

the  other  functions  of  15°. 


find 


sin  15° 
cos  15° 


=  2 -Vs. 


sin2 15°  +  cos2 15°  =  1. 
sin  15°  =  cos  15°  (2 -VS), 
[cos(2-V3)]2  +  cos2  =  l, 
cos2  (4  -  4V3  +  3)  +  cos2  =  1, 
cos2(8-4V3)  =  l. 

1  2+V3 


cos2 15°  = 


4(2 -Vs)  4 

cos  i5°  =  y2+V3^,y^;;^ 

sin^  =  1  —  cos^. 

sinn5°  =  l-2iV3^2-V3 
4  4 


in  15°  =  aV2-V3. 


sin 


cot  15°  = 


tan  15°      2-V3 
=  2+V3. 


17.   Given  cot  22°  30^=  V2  +  1 ; 
find  the  other  functions  of  22°  SO''. 


tan  =  —  = 


1 


cot      V2  +  1 


=V2-1. 


sm       , 
—  ==  tan, 
cos 

cos^  +  sin^  =  1. 


(1) 

(2) 


From  (]),  cos  tan  =  sin. 

Squaring,  cos^tan^  =  sin'^ 
From  (2),  cos^        =  —  sin^  +  1 
Add,  cos^tan^  +  cos^  =  1 

cos^(tan2  +  1)  =  1, 
cos2(4-2V2)  =  l, 

cosV4-2V2  =  l. 


18 


TRIGONOMETRY. 


.*.  COS  = 


V4-2V2 


=v 


4  +  2\/2 


=  aV2+V2. 


sm 


=  ^1_2±V2 


\ 


'4-2-V2 


=.^'2z:^=iV2:vI. 

18.   Given  siB(F  =  0;    find    the 
other  fanctions  of  0°. 


cos  =  Vl  —  sm^  =  Vl  -0. 
.*.  cos  =  1. 

tan=?^=?  =  0. 
cos       1 

cot  = =  -  =  00. 

tan     0 

1        1      1 
sec  =  —  =  -  =  1. 

cos      1 
1       1 

CSC  = =  _  =  00. 

sin     0 

19.   Given   sin  90°=  1;   find  the 
other  functions  of  90°. 

sin  90°  =  1. 

cos  =  Vl  —  sin^  =  0. 

.  sin      1      _ 

tan  =  — -  =  -  =  CO. 

cos      0 

cot  =  —  =  —  =  0. 

tan      CO 

1       1 

sec  =  —  =  -  =  CO. 

COS      0 
1       1      1 

CSC=  -r-  =  7=  i. 

sm      1 


20,   Given  tan  90°  =  00  ;  find  the 
other  functions  of  90°. 


tan  90°=  00. 
tan      00 


cot  =  —  =  —  =  0. 


sm 
cos 


sin^  =  CO  cos^ 
sin^  +  cos^  =  1 


—  COS^  =  CO  cos^  —  1 
CO  cos^  =  1. 


\  CO 


COS  =  -xi —  =  0. 

sin 

=  00. 

0 
•     sin  =  1. 

1 

sec  =  -  =  00. 
0 

CSC    =  1. 

21.   Express  the  values  of  all  the 
other  fanctions  in  terms  of  sin  A. 

By  formnlae  on  pages  11  and  12, 
sin  A  =  sin  A, 


cos  A  =  Vl  —  sin^^, 

,        .  sin  ^ 

tan^  = 


cot  J.= 
sec  A  = 
CSC  A  = 


Vl  —  sin'^^l 

Vl  -  sin^^ 

sin  A 

1 

Vl  -  sinM 

1 
sin^ 


22.   Express  the  values  of  all  the 
other  fanctions  in  terms  of  cos  A. 


TEACHEE-S     EDITION. 


19 


By  formula  on  pages  11  and  12, 


sin  A  = 

Vi- 

COS 

'A 

cos  A  = 

ces  A, 

tan  J.= 

Vi- 

COS 

'A 

cos  A 

cotA  = 

COf 

A 

1 

Vi- 

cos 

'A 

sec  A  = 

1 

COS  J. 

, 

CSC  A  = 

1 

' 

Vi- 

cos 

'A 

23.    Express  the  values  of  all  the 
other  functions  in  terms  of  tan  J.. 

cot^ 


tan^ 

-  =  tan  A. 
b 

a?  +  h''  =  1. 

a  =  h  tan  A. 

a?  =  6^  tan^  J.. 

a2-62tan2^   =0 
a2  +  62  =1 

h-'  (i  +  tan^.l)  =  1 

1  +  tan^^ 
sin  A  =  Vi  —  cos'^^ 


V 


=  Al- 


1  +  tan^^ 


4 


1  -  1  +  tan^J. 
1  +  tan2  J. 

tan  J. 


Vi  +  tan^  J. 


sec  A  =  —  =  VrTtanlJ. 
cos 


4       1        Vi  +  tan'-^J. 
CSC  ^  =  —  = 


sm 


tan^l 


24.    Express  the  values  of  all  the 
other  functions  in  terms  of  cot  A. 

=  tan  A. 


cot^ 
sin  A 


=  tan  A. 


cos  A 
Let  X  =  sin,     y  =  cos. 

x__l 

y     cot  A 
X  cot  A  =  y, 
x^  cot^A  =  y"^. 

x^co\?A-y'^  =  ^ 
x^  +3/^  =  1 


a;2(l+cot2J.)  =  l 
1 


1  +  C0t2  J. 


sin  A  = 


cos 


Vi  +  cot^^ 
^  =  Vl  —  sin'-^J. 


V 


=  Al- 


=v 


1  +  cot^J. 
1  +  cot^A  -  1 

1  +  C0t2  J. 

cot  J. 


VTTcoFl 


sec  A  = 


cos -4 


CSC 


\/\  +  COt''^  J. 

cot^ 

A  =  ^^  =  VlTcot^Z. 
sin  A 


25.   Given  2  sin  ^  =  cos  A  ;   find 
sin  A  and  cos  A. 


20 


TRIGONOMETRY. 


sin^J.  +  cos^J.  =  1. 
sin^  J.  +  4  sin^  J.  =  1. 
5  siu^A  =  1, 


sinM  =  i- 
5 


sin  A 


'  n 


iV5. 


".  cos  J.  =  |V5. 


26.   Given  4  sin  ^  =  tan  ^  ;    find 
sin  A  and  tan  J.. 

J,      sin  ^ 


But 


cos  A 

tan  J.= 

=  4  sin  A. 

4  sin  J.  = 

sin  A 
cos  A 

4  sin  A ; 

K  cos  A  =  sin 

^. 

cos  A 

sin  ^       1 
4sinA     4 

sin^J. 

+  cos^J.  =  1. 

sin  A 

>'16 

tan  J. 

sin  A 
cos  A 

15 

27.   If  sin  ^  :  cos  J.  =  9  :  40,  find 
sin  A  and  cos  A. 

40  sin  J.  =  9  cos  A. 
(sq.)  1600  sin^A  -  81  cosM. 
1600  sin2^  -  81  cosM  -  0. 
But      sin2^  +  cos2  J.  =  1. 


Multiplying  bj 

'  81  and  adding, 

1681  sin 

2^1  = 

=  81, 

.'.  41  sin 

A  = 

9. 

sin 

A  = 

_  9 
'4l" 

sin^J. 

+  cc 
A== 

A  = 

)S2A  = 

=  1. 

cos 

Vi- 

-  smM. 

.:.  cos 

v^- 

{  ^  Y 

40 
'41 

28.  Transform  the  quantity  tan^  J. 
+  cot^-^  —  sin^J.  —  cos'^J.  into  a  form 
containing  only  cos  A. 


idX^A  = 
coi^A  = 


sinM  _  1  —  cos^^ 
cos^A         cos'-^J. 

cos^J.         cos^^ 


sin^A      1  —  cos^^ 
cos2^ 


1  —  cos^tI 

cos^  J.  1  —  COS^A 

—  1  +  cos^yl  —  COS^J. 
1—  2  cos^  J-+2  COS*  J.— cosM+cos^^ 

COS^^  —  COS*  J. 

_  1  -  3  cos^^  +  3  cos*  J. 
cos^  A  —  cos*^ 

29.  Prove  that  sin  J.  +  cos  J.  = 

(1  +  tan  A)  cos  A. 

^i5_4=.tanA 
cos  A 

sin  A  =  tan  J.  cos  ^. 

sin  A  +  cos  A  =  tan  ^  cos  J.  +  cos  A 
=  (1  +  tan  A)  cos  A. 

30.  Prove  that  tan^  +  cot^  = 
sec  A  X  CSC  A. 


TEACHERS     EDITION. 


21 


tan  A 
cot  J.= 


_  sin  A 
cos  A 
cos  A 
sin  A 


tan^  +  cot^=^HL^  +  ^-^^ 
cos  A     sin  A 


EXEECISE   VI. 

1,  In  Case  II.  give  another  way 
of  finding  c,  after  h  has  been  found. 

cos  A  =  -, 
c 

h  =  c  cos  A, 
h 

c  =  ■ 

COS  A 

2.  In  Case  III.  give  another  way 
of  finding  c,  after  a  has  been  found. 

■        A         « 

c 

c&va.  A  =  a, 

a 
c  = 


sin  A 

3.  In  Case  IV.  give  another  way 
of  finding  6,  after  the  angles  have 
been  found. 

cos  A  =  -y 

c 
6  =  c  cos  A. 

4.  In  Case  V.  give  another  way 
of  finding  c,  after  the  angles  have 
been  found. 

A       ^ 
cos  A  =  —, 

G 


c  cos  J.  =  h, 


c  = 


cos  J. 


sin^^  +  cos^^ 
cos  A  sin  A 


^va?A  +  cos^J.  =  1. 
tan  J.  +  cot  A- 


cos 

^sin^ 

=  sec 

A  Xcsc 

A 

Page 

20. 

5 

.   Given 
sir 

B 
A 

vA 

a 

and  c; 

=  (90° 
a 
c 

=  c  sin 

find  A, 
A, 

a, 

h 

cos -4 

b 
c 

h 

=  C  COS 

A. 

6.   Given  B  and  h  ;  find  A,  a,  c. 
A  =  (90°  -  B), 


tan  J.= 

a 
h 

a  = 

h  tan^ 

sin  B  = 

h 
— 1 
c 

b  = 

c  sin  B^ 

c  = 

h 

sin  5 

7.  Given  B  and  a ;  find  A,  b,  c. 
A  =  (90°  -  B), 
tan  J.  =  -) 


b  tan  A  =  a, 


tan  J. 


22 


TRIGONOMETEY. 


sin  A 


c  &'m  A  =  a, 


c  = 


sin  J. 

8.   Given  b  and  c ;  find  A,  b,  a. 
cos  J.  =  -. 


B  = 

=  (90° 

-A). 

sin 

A  = 

a 
c 

a  = 

=  c  sin 

A. 

9.   Given  a  =  6,  c  =  12;  required 
^  =  30°,  B  =  60°,  b  =  10.392. 


sin 

A 
A 

a     1       • 

=  -  =  -  =  sm 

c      2 
=  30°. 

30°, 

£ 

=  (90°  -A)  = 

=  60°. 

cos 

A 
b 

b 

~  c 

=  c  cos  A. 

log 

cos 

A 

=  9.93753 

log 

12 

=  1.07918 

log 

b 

=  1.01671 

b 

=  10.392. 

10.   Given  ^  =  60°,    6  =  4;     re- 
quired B  =  30°,  c  =  8,  a  =  6.9282. 

Since  A  =  60°  and  6  =  4, 

B  =  (90°  -  60°)  =  30°, 
and  c  =  8.     (By  Geometry.) 

c^  =  a^  +  b^. 
.•.c2-62  =  a2  =  48. 


log48  =  loga2=  1.68124, 
log  a  =  0.84062, 
a  =  6.9282. 

11.  Given  ^  =  30°,  a  =  3;  re- 
quired B  =  60°,  c  =  6,  6  =  5.1961. 

Since  A  =  30°  and  a  =  3, 

B  =  (90°  -  30°)  =  60°, 
and  c  =  Q. 

c'-  =  a?  -\-  b^. 
.-.  c2-a2  =  62  =  27. 

log27  =  log  62  =  1.43136, 
log  6  =0.71568, 
6  =5.1961. 

12.  Given  a  =  4,  6  =  4;  required 
J.  =  ^=45°,  c  =  5.6568. 

Since  a  and  6  each  =  4,  the  A  is 
an  isosceles  A,  and  the  A  A  and  B 
are  equal. 

:.A  =  ^  of  90°  =  45°, 

5  =  J  of  90°  =  45°. 

c2  =  a2  +  62  =  32. 

log  32  =  log  c2=  1.50515, 

log  c  =  0.75257, 

c  =5.6568. 

13.  Given  a  =  2,  c  =  2.82843  ; 
required  A=^B  =  45°,  6  =  2. 


6  =  \/^ 


=  ■\J{c-[-a){c  —  a), 
log  62  =  log  {c  +  a)  +  log  (c—  a). 

log  {c^a)=  0.68381 

log  (c  -  a)  =  9.91826-10 

log  62        -  =  0.60207 


TEACHEES     EDITION". 


23 


log  h  =  0.30103, 

h  =2. 

.*.  the  A  is  an  isosceles  rt.  A. 
.-.  ^  =  ^  =  45°. 

14.  Given  c  =  627,  A  =  23^  30^ ; 
required  B  =  66°  30^  a  =  250.02, 
b  =  575.0. 

B  =  (90°  -  ^)  =  66""  30^. 

a  =  c  sm  A. 

log  a  =  log  c  +  log  sin  A. 

log  c  =  2.79727 
log  sin  A  =  9.60070 
log  a  =  2.39797 
a  =  250.02. 

6  =  c  cos  A. 
log  6  =  log  c  +  log  cos  A. 

log  c  -  2.79727 
log  cos  A  =  9.96240 
log  6  =2.75967 
h  =  575. 

15.  Given  c  =  2280,  A  =  28°  5^ ; 
required  B  =  61°  55^,  a  =  1073.3, 
6  =  2011.6. 

B  =  61°  55^ 
a  =  c  sin  A. 
log  «  =  log  c  +  log  sin  ^. 

log  c  =  3.35793 
log  sin  A  =  9.67280 
log  a  =  3.03073 
a  =  1073.3. 

5  =  c  cos  A. 
log  5  =  log  c  +  log  cos  A. 


log  c  =  3.35793 
log  cos  A  =  9.94560 
log  b  =  3.30353 
b  =2011.6 

16.  Given  c  =  72. 15,  A  =  39°  34^; 
required  B  =  50°  26^  a  =  45.958, 
b  =  55.620. 

^  =  50°  26^ 
a  =  c  sin  J., 
log  a  =  log  c  +  log  sin  A, 

log  c  =  1.85824 
log  sin  A  =  9.80412 
log  a  =  1.66236 
a  =  45.958. 

6  =  c  cos  A. 
log  6  =  log  c  +  log  cos  A. 

log  c  =  1.85824 
log  cos  A  =  9.88699 
log  b  =  1.74523 
b  =  55.620. 

17.  Given  g=-1,  A  =  36°;  required 
5  =  54°,  a  =  0.58779,  5  =  0.80902. 


B 

=  54°. 

sin 

A 
a 

_  a 
c 
=  c  sin  A. 

log  a 

=  log  c  +  log 

sin  A 

log 

c 

=  0.00000 

log 

sin 

A 

=  9.76922 

log 

a 

=  9.76922 

a 

=  0.58779. 

cos 

A- 

c 

24 


TRIGONOMETRY. 


b  =  c  cos  A. 
log  6  =  log  c  +  log  cos  A. 

log  c         =  0.00000 
log  cos  A  =  9.90796 
log  b         =  9.90796  -  10 
b  =  0.80902. 


18.  Given  c  =  200,  B  =  21°  47^ ; 
required  ^  =  68°  13^  a  =  185.72, 
b  =  74.219. 


13^ 


sm  -d  =  — 
c 


a  =  c  sin  ^. 
log  a  =  log  c  +  log  sin  A. 

log  c  =  2.30103 
log  sin  A  =  9.96783 
log  a  =  2.26886 
a  =  185.73. 

cos  .4  =  -• 
c 

&  =  c  COS  .4. 

log  6  =  log  c  +  log  COS  A. 

log  G  =  2.30103 
log  cos  A  =  9.56949 
log  b  =  1.87052 
6  =  74.22. 

19.  Given  c  =  93.4,  B  =  76°  25^  ; 
required  A  =  13°  35^  a  =  21.936, 
6  =  90.788. 

A  =  13°  35^ 
a  =  c  sin  A. 
log  a  =  log  c  +  log  sin  A. 


logc 

= 

1.97035 

log  sin 

A  = 

9.37081 

log  a 

= 

1.34116 

a 

= 

21.936. 

b  = 

a  cot  A. 

log 

b  = 

log  a  +  log 

cot  A 

log  a 

= 

=  1.34116 

log  cot  A  = 

=  0.61687 

log  b 

= 

=  1.95803 

b 

= 

=  90.788. 

20.   Given 

a  =  637,   A 

=  4°  35^  ; 

required 

B  = 

85°  25^  b  = 

7946,  c  = 

7971.5. 

B^ 

=  85°  25^ 

b  = 

=  a  cot  A. 

lo 

gb- 

=  log  a  +  log 

cot  A. 

log  a 

= 

=  2.80414 

log  cot  A  = 

=  1.09601 

log  b 

= 

=  3.90015 

b 

= 

=  7946. 

logc 

= 

=  log  a  +  colog  sin  A. 

log  a 

= 

=  2.80414 

colog  sin 

A  = 

=  1.09740 

log  c         =  3.90154 
c  =  7971.5. 

21.  Given  a  =  48.532,  ^  =  36°  44^; 
required  B  =  53°  16^  b  =  65.033, 
c  =  81.144. 

^  =  90°  -  .4 

=  90°  -  36°  44^ 

=  53°  16^ 

•        A        « 

sm  A  =  — 
c 


TEACHERS     EDITION. 


25 


c 

sin  J. 

logc 

=  log  a  +  colog 

sin 

log 

a 

=  1.68603 

colog 

sin  A 

=  0.22323 

log 

c 

=  1.90926 

c 

=  81.144. 

cos  A 

b 

b  =  c  cos  A. 
log  b  =  log  c  +  log  cos  A. 

log  c  =  1.90926 
log  cos  A  =  9.90386 
log  b  =  1.81312 
b  =  65.031. 

22.  Given  a  =  0.0008,  ^  =  86°; 
required  B  =  ^°,b  =  0.0000559,  c  = 
0.000802. 

B  =  dO°-A 

=  90°  -  86° 

=  4°. 

sin^  =  -- 


sin  J. 
log  c  =  log  a  +  colog  sin  A. 

log  a         =  6.90309  -  10 
colog  sin  A  =  0.00106 
log  c         =  6.90415  -  10 
c  =  0.000802. 

cos  A  =  ~ 
c 

b  =  c  cos  A. 

log  6  =  log  c  +  log  cos  A. 


log  c         =  6.90415  -  10 
log  cos  A  =  8.84358 
log  b        =  5.74773  -  10 
b  =  0.0000559. 

23.  Given  6  =  50.937,  5  =  43°  48^; 
required  ^  =  46°  12^  a  =  53.116, 
c  =  73.59. 

A  =  46°  12^ 

tan^  =  -. 
b 

a  =  b  tan  A. 

log  5         =  1.70703 
log  tan  A  =  0.01820 


log 

a 

=  1.72523 

a 

=  53.116. 

sin 

A 

a 
c 
a 

sin  A 

log  a        =  1.72523 

colog  sin  A  =  0.14161 

log  c         =  1.86684 

c  =  73.593. 

24.  Given  5  =  2,  5  =  3°  38^  ;  re- 
quired A  =  86°  22^,  a  =  31.497,  c  = 
31.560. 

A  =  86°  22^ 

tan  .4  =  -. 
6 

a  =  5  tan  A. 
log  6  =  0.30103 
log  tan^  =  1.19723 
log  a  =  1.49826 
a  =  31.496. 


26 


TRIGONOMETRY. 


sin 

A 

a 
c 

c 

a 

sin  J. 

log 

a 

=  1.49826 

colog 

sin 

A 

=  0.00087 

log 

c 

=  1.49913 

G 

=  31.560. 

log  a         =  1.86332 
colog  sin  A  =  0.44305 


25.   Given  a  =  992,  B  =  76°  19^ ; 
A  =  13°  41^  b  =  4074.45,  c  =  4193.55. 

^  =  90°  -  76°  19^ 
=  13°  41^ 

sm  A  =  — 
c 

log  c  =  log  a  +  eolog  sin  A. 

log  a        =  2.99651 

colog  sin  A  =  0.62607 

log  c         =  3.62258 

c  =  4193.6. 

sin  B  =  — 
c 

log  6  =  log  c  +  log  sin  B. 

log  c  =  3.62258 
log  sin  B  =  9.98750 
log  h  =  3.61008 
h  =  4074.5. 


26.   Given   a  =  73,    5  =  68°  52^ ; 

required  A  =  21°  8^  h  =  188.86,  c  = 

202.47. 

^  =  90°  -  .5  =  21°  8^ 

•        A         ^ 

sm  A  =  — 

c 

log  c   =  log  a  +  colog  sin  A. 


log 

c 

= 

'-  2.30637 

c 

= 

=  202.47. 

sin 

B  = 

6 
c 

log 

h  = 

=  log  c  +  log  sin 

B. 

log 

c 

= 

=  2.30637 

log 

sin 

B  = 

=  9.96976 

log 

5 

= 

=  2.27613 

& 

= 

=  188.86. 

27. 

Given 

a  =  2.189,  B  = 

45°  25^; 

required 

yl 

=  44°  35^     h  = 

2.2211, 

c  =  3.1185. 

^  = 

=  90°  -  45°  25^ 
=  44°  35^ 

sin 

A  = 

c- 

a 
c 
a 

sin^ 

lo^ 

yc- 

=  log  a  +  colog 

sin  A. 

log 

a 

= 

=  0.34025 

colog 

sin 

^  = 

=  0.15370 

log 

G 

: 

=  0.49395 

c 

=  3.1185. 

cos 

A. 
h- 

b 

c 

=  c  cos  A. 

lo^ 

l^ 

=  log  c  +  log  cos  A. 

log 

c 

=  0.49395 

log 

cos 

A 

=  9.85262 

log  6 
b 


=  0.34657 
=  2.2211. 


28.   Given  6  =4,  ^  =  37°56^  re- 
quired  5=52°4^  a  =3.1176,   c  = 

5.0714. 


teachers'  edition. 


27 


B  =  52°  ¥. 
h 


cos  A  = 
c 

h  =  c  cos  ^. 

h 
c  = 

COS  J. 
log  G  =  log  h  +  Colog  COS  A. 

log  5         =  0.60206 
colog  COS  A  =  0.10307 

log  c  =  0.70513 
c  =  5.0714. 

tan  A  =  -- 

0 

a  =  b  tan  ^. 
log  a  =  log  &  +  log  tan  A. 

log  h  =  0.60206 
log  tan  A  =  9.89177 
log  a  =  0.49383 
a  =  3.1176. 

29.   Given    c  =  8590,    a  =  4476 
required  ^ 
6  =  7332.8. 


31°  24^  B  =  58°  36^ 


Sin  A  =  — 


log  sin  J.  =  log  a  +  colog  c. 

log  a        =  3.65089 
colog  c         =  6.06601 

log  sin  A  =  9.71690 
A  =  31°  24^. 

B  =  58°  36^ 

cot  A  =  — 
a 

log  h  =  log  a  +  log  cot  A. 

log  a        =    3.65089 
log  cot  A  =  10.21438 

log  b        =    3.86527 
b  =  7332.8. 


30.  Given  c  =  86.53,  a  =  71.78  ; 
required  A  =  56°  3^  B  =  33°  57^ 
b  =  48.324. 

log  sin  A  =  log  a  +  colog  c. 
log  a        =  1.85600 
colog  c  =  8.06283 

log  sin  A  =  9.91883 
A  =  56°  3^ 

.5  =  33°  57^ 

log  6  =  log  a  +  log  cot  A. 
log  a         =  1.85600 
log  cot  J.  =  9.82817 
log  b         =  1.68417 
&  =  48.324. 

31.  Given  c  =  9.35,  a  =  8.49  ; 
required  A  =  65°  14^  B  =  24°  46^ 
6  =  3.917. 

A  <^ 

sm  J.  =  — 
c 

.5  =  90°  -  .4. 

colog  c         =  9.02919 

log  a        =  0.92891 

log  sin  A  =  9.95810 
A  =  65°  14^ 

B  =  24°  46^ 

cos  A  =  — 
c 

6  =  c  COS  A. 

log  c  =  0.97081 
log  cos  A  =  9.62214 
log  b  =  0.59295 
b  =  3.917. 

32.  Given  c=2194,  6  =  1312.7; 
required  ^=53°  15',  B  =  36°  45^ 
a  =  1758. 


28 


TRIGONOMETRY. 


COS  A  = 


log  6 

=  3.11816 

colog  c 

=  6.65876 

log  COS 

A  =  9.77692 

A 

=  53° 15^ 

B 

=  (90°  -  A) 

=  36°  W, 

Sin  A  =  -' 
c 

a  =  c  sin  A. 

log  c         =  3.34124 

log  sin  A  =  9.90377 

log  a        =  3.24501 

a  =  1758. 

33.  Given  c  =  30.69,  b  =  18.256  ; 
required  A  =  53°  30^  B  =  36°  30^ 
a  =  24.67. 

A         ^ 

cos  A  =  — 
c 

log  cos  J.  =  log  b  +  colog  c. 

log  6        =  1.26140 
colog  c         =  8.51300 

log  cos  ^  =  9.77440 
A  =  53°  30^. 

£  =  36°  30^ 

tan  A  =  -- 

0 

log  a  =  log  tan  A  +  log  6. 

log  tanJ.  =  10.13079 
log  b  =  1.26140 
log  a  =  1.39219 
a  =  24.671. 

34.  Given  a  =  38.313,  b  =  19.522 ; 
required  A  =  63°,  .B  =  27°,  c  =  43. 


tan  A  =  -- 
b 

log  tan  A  =  log  a  +  colog  h. 
log  a        =    1.58335 
colog  b         =    8.70948 


log 

tan  J. 

=  10.29283 

A 

=  63°. 

B 

=  27°. 

sin -4 

a 

logc 

=  log  a  +  colog  sin 

A 

log 

a 

=  1.58335 

colog 

sin  A 

=  0.05012 

log 

c 

=  1.63347 

c 

=  43. 

35.  Given  a  =  1.2291,  6  =  14.950; 
required  J.  =  4°  42^  B  =  85°  18^, 
c  =  15. 

tan^  =  ?. 
b 

log  a        =  0.08959 

colog  b         =  8.82536  - 10 

log  tan^  =  8.91495 

A  =   4°42^ 

B'  =  85°  18'. 


sin  A  = 


a 


a  =  c  sin  A. 
a 
sm-d 

log  a        =  0.08959 
colog  sin  A  =  1.08651 

log  G         =  1.17610 
c  -15. 


TEACHERS     EDITION. 


29 


36.  Given  a  =  415.38,  &  =  62.080; 
required  ^  =  81°  30^  .S  =  8°  30^ 
c  =  420. 

tan  A  =  -■ 

0 

log  a        =   2.61845 
colog  b         =    8.20705  -  10 
log  tan  A  =  10.82550 
A  =  81°  30^ 

B  =   8°30^ 

•  A  ^ 

Bin  A  =  -- 


a 

=  c  sin  A. 

c 

a 

sin  J. 

log  a 

=  2.61845 

colog  sin 

A 

=  0.00480 

logc 

=  2.62325 

c 

=  420. 

37.  Given  a  =  13.690,  6  =  16.926; 
required  A  =  38°  58^  B  =  51°  2', 
c  =  21.77. 

tan  A  =  -' 

0 

log  tan  A  =  log  a  +  colog  b. 

log  a        =  1.13640 
colog  b        =  8.77144  - 10 
log  tan7l  =  9.90784 
A  =  38°  58^ 

B  =51°    2\ 


A  Of 


c  = 


sin  J. 
log  c  =  log  a  +  colog  sin  A. 


log  a        =  1.13640 
colog  sin  A  =  0.20144 
log  c         =  1.33784 
c  =  21,769. 

38.  Given  e  =  91.92,  a  =  2.19; 
required  J.  =  1°  21^  55^^  ^  =  88° 
38^  5^^  b  =  91.894. 

•        A         ^ 

sin  ^  =  — 
c 

log  sin  -4  =  log  a  +  colog  c. 

log  a        =  0.34044 
colog  c         =  8.03659  -  10      " 

log  sin  ^  =  8.37703 

A  =1°  21^  55^^. 

B  =88°  38^    5^^ 

cos  A  =  -' 
c 

6  =  c  cos  -4. 
log  b  =  log  c  +  log  cos  A. 

log  c         =  1.96341 
log  cos  A  =  9.99988 

log  6        =  1.96329 
b  =  91.894. 

39.  Compute  the  unknown  parts 
and  also  the  area,  having  given 
a  =  5,  6  =  6. 

tan  A  =  — 
b 

log  tan  A  =  log  a  +  colog  6. 


log  a 
colog  6 


=  0.69897 

=  9.22185-10 


log  tan  J.  =  9.92082 
A  =  39°  48^ 

B  ^  50°  12'. 


TRIGONOMETRY. 


sin  A  =  — 
c 


sin  A 
log  c  =  log  a  +  colog  sin  A. 

log  a        =  0.69897 
colog  sin  A  =  0.19375 
log  c         =  0.89272 
c  =  7.8112. 

axh      30 


F 


15. 


40.  Compute  the  unknown  parts 
and  also  the  area,  having  given 
a  =  0.615,  c  =  70. 

I      .      .      a 
sm  A^-- 

G 

log  sin  J.  =  log  a  +  colog  c. 
log  a        =9.78888-10 
colog  c         =  8.15490-10 
log  sin  A  =  7.94378 
A  =  30^  12^^. 

£  =  89°  29^  48'^ 

tan  A^-- 

0 

log  &  =  log  a  +  colog  tan  A. 
log  a        =  9.78888  - 10 

colog  tan  A  =  2.05626  

log  b         =  1.84514 
b  =  70.007. 


F= 

--lab. 

log« 

=  9.78888- 

-10 

log  6 

=  1.84514 

colog  2 

=  9.69897  - 

-10 

41.   Compute  the  unknown  parts 
and   also   the  area,   having    given 


b 

=  ^2. 

c=\/3. 

n/3  = 

cos^  = 

1.25991. 

1.73205. 

b 

c 

log 

COS  A  = 

--  log  b  +  colog  c. 

log 

b 

=  0.10034 

colog 

c          = 

9.76144- 

-10 

log  F       =  1.33291 
F  =  21.528. 


log  cos  A  =  9.86178 
A  =  43°  20^ 

B  -  46°  40^ 

.      a 
sm  A  =  — 
c 

a  =  c  sin  A. 
log  a  —  log  c  +  log  sin  A. 

log  c         =  0.23856 
log  sin  A  =  9.83648 

log  a        =  0.07504 
a  =  1.1886. 

J'^  J  ab. 

log  a  =  0.07504 

log  b  =  0.10034 

colog  2  =  9.69897  - 10 

log  F  =  9.87435  -  10 

F  =  0.74876. 

42.  Compute  the  unknown  parts 
and  also  the  area,  having  given 
a  =7,  il  =  18°14^ 

sm  A  =  — 
c 


c  = 


sin  A 


TEACHERS     EDITION. 


31 


log  c  =  log  a  +  colog  sin  A. 

log  a         =  0.84510 
colog  sin  A  =  0.50461 

log  c         =  1.34971 
c  =  22.372. 


tan  J.  =  -• 


tan  J. 
log  h  =  log  a  +  colog  tan  ^. 

log  a        -  0.84510 
colog  tan  A  =  0.48224  -  10 

log  b        =  1.32734 
b  =  21.249. 


log  a 

log  6 

colog  2 

logi^ 


F=^ab. 

=  0.84510 
=  1.32734 
=  9.69897-10 
=  1.87141         " 
■     =  74.371. 


43.  Compute  the  unknown  parts 
and  also  the  area,  having  given 
6  =  12,  ^=29°8^ 

A  =  29°  8^. 
B  =  60°  52^ 

COS  A  =  — 
c 

h 
c  = 


COS  A 
log  c  =  log  b  +  colog  COS  A. 

log  b         =  1.07918 

colog  COS  A  =  0.05874 

log  c         =  1.13792 

c  =  13.738. 


sm  ^  =  -. 
c 

a  =  c  sin^. 
log  a  =  log  c  +  log  sin  A. 

log  c  =  1.13792 
log  sin  A  =  9.68739 
log  a  =  0.82531 
a  =  6.6882. 

F=  I  ab. 

log  F=  log  a  +  log  5  + colog  2. 

log  a  =  0.82531 
=  1.07918 
=  9.69897-10 


log  5 
colog  2 
logF 

F 


=  1.60346 
=  40.129. 


44.  Compute  the  unknown  parts 
and  also  the  area,  having  given 
c  =  68,  ^  =  69°54^ 

A  =  69°  54^ 

B  =  20°  6^ 

sm  A  =  -' 
c 

a  =  c  sin  A. 
log  a  =  log  c  +  log  sin  A. 

log  c         =  1.83251 
log  sin  A  =  9.97271 

log  a         =1.80522 
a  =  63.859. 


cos  A  =  -. 
c 

6  =  c  cos  ^. 
log  6  =  log  c  +  log  cos  A. 


32 


TRIGONOMETKY. 


log  c  =  1.83251 
log  cos  A  =  9.53613 
log  b  =  1.36864 
b  =  23.369. 

F=  J  ab. 

log  a  =  1.80522 

log  b  =  1.36864 

colog  2  =  9.69897-10 

log  i^  =  2.87283 

F  =746.15. 

45.  Compute  the  unknown  parts 
and  also  the  area,  having  given 
c  =  27,  B  =  M°4:'. 

A  =  45°  56^ 

a  =  c  sin  A. 

f  log  a  =  log  c  +  log  sin  A. 

log  c  =  1.43136 
log  sin  A  =  9.85645 
log  a  =  1.28781 
a  =  19.40. 

6  =  c  cos  A. 
log  5  =  log  c  +  log  cos  A. 

log  c  =  1.43136 
log  cos  A  =  9.84229 
log  b  =  1.27365 
&  =  18.778. 

F=  J  a6. 

log  a  =  1.28781 

log  b  =  1.27365 

colog  2  =9.69897-10 

log  F  =  2.26043 

-F  =  182.15. 


46.  Compute  the  unknown  parts 
and  also  the  area,  having  given 
a  =  47,  ^  =  48°  49^. 

A  =  41°  IK 

b  =  a  cot -4, 
log  b  =  log  a  +  log  cot  A. 

log  a  =  1.67210 
log  cot  A  =  10.05803 
log&  =  1.73013 
b  =  53.719. 


&iuA 

log  c  =  log  a  +  colog  sin  A. 

log  a 

=  1.67210 

colog  sin 

^  =  0.18146 

logc 

=  1.85356 

c 

=  71.377. 

F=lab. 

log  a 

=  1.67210  . 

log  6 

=  1.73013 

colog  2 

=  9.69897-10 

logF 

=  3.10120 

F 

=  1262.4. 

47.   Compute  the  unknown  parts 
and   also   the   area,   having   given 

6  =  9,  .5 

=  34°  44^ 

A  =  55° 16^ 

a  =  b  tan  A. 

log 

a  =  log  b  +  log  tan  A. 

log  6 

=    0.95424 

log  tan  J.  =  10.15908 

log  a 

=    1.11332 

a 

=  12.98L 

TEACHERS     EDITION. 


33 


c  = 


mi  A 
log  c  =  log  a  4-  colog  sin  A. 

log  a         =1.11332 
colog  sin  ^"==0.08523 

log  c  =  1.19855 

c  =  15.7960 


log  a 

log  6 
colog  2 

logi^ 


i^-  ^  ah. 

=  1.11332 
=  0.95424 
=  9.69897-10 

=  1.76653 
=  58.416. 


48.  Compute  the  unknown  parts 
and  also  the  area,  having  given 
c  =  8.462,  B  =  m°¥. 

^  =  3°  56^ 
a  =  c  sin  A. 
log  a  =  log  c  +  log  sin  A. 

log  c         =  0.92747 
log  sin  A  =  8.83630 

log  a         =9.76377-10 
a  =  0.58046. 

h  =  c  cos  A. 
log  h  =  log  c  +log  cos  ^. 
log  c         =  0.92747 
log  cos  A  =  9.99898 

log  b         =  0.92645 
6  =  8.442. 


F 

=  lah. 

log  a 

=  9.76377- 

-10 

log& 

=  0.92645 

colog  2 

=  9.69897  - 

-10 

logi^ 

=  0.38919 

F 

=  2.4501. 

49.  Find  the  value  of  F  in  terms 
of  c  and  A. 

F=iab. 

Sin  A  =-' 
c 

a  =  c  sin  A. 

COS  A  =  — 
c 

b  =  c  COS  A. 

Substitute, 

F=^ab 

=  J  (c^  sin  A  COS  A). 

50.  Find  the  value  of  F  in  terms 
of  a  and  A. 

F=  J  ab. 

cot  A  =  — 

a 

b  =  a  cot  A. 
Substitute, 

F=^ab 

=  ^  {a^  cot  ^). 

51.  Find  the  value  of  F  in  terms 
of  b  and  A. 

F=lab. 
tan  ^  =  -• 

a  =  b  tan  .4. 

Substitute, 

i^=Ja& 

=  J  (62  tan  ^). 

52.  Find  the  value  of  F  in  terms 
of  a  and  c. 

F=  i  ab. 
c^  =  a^  +  b^. 
b^  =  c^  —  a^, 

b  =  a/c^  —  a^. 

Substitute, 


34 


TRIGONOMETRY. 


53.   Given  i^=  58,  a  =  10  ;  solve 
the  triangle. 

F=  I  ah. 

a 

log  6  =  log  2  i^  +  colog  a. 

log2P  =2.06446 

colog  a  =  9.00000  -  10 

log  h  =  1.06446 

h  =  11.6. 

tan  A  =  -■ 

0 

log  tan  A  =  loga  +  colog  b. 

log  a        =  1.00000 
colog  b         =  8.93554  -  10 
log  tan^  =  9.93554 
A  =  40°  45^  48^^ 

B  =49°14M2^>'. 


sin^ 
log  €  =  log  a  +  colog  sin  A. 

log  a        =  1.00000 

colog  sin  .4  =  0.18513 

log  c         =  1.18513 

c  =  15.315. 

54.   Given  i^=  18,   6  =  5;   solve 
the  triangle. 

F=  I  ah. 
2F 

log  a  =  log  2  i^  +  colog  6. 

log2i?'     =1.55630 
colog  b         =  9.30103  -  10 

log  a        =  0.85733        " 
a  =  7.2. 


tan  A  =  -' 

0 

log  tan  A   =  log  a  +  colog  b. 

log  a  =    0.85733 

colog  6  =    9.30103 -10 


log 

tan  J.    = 

10.15836 

A 

55°  13^  20^^. 

B 

34°  46^  40^^ 

c  = 

a 

sin^ 

logc  = 

=  log  a  +  colog  sin  A. 

log 

a          = 

0.85733 

colog 

sin  A    = 

0.08546 

log 

c           = 

0.94279 

c 

8.7658. 

55. 

Given  jP=  12,  ^  =  29° 

solve 

the  triangle. 

B  = 

=  61°. 

■  F= 

-lab  =  12. 

ah  = 

=  24. 

a  = 

24 
■  b' 

tan^  = 

a 

tan  29°- 

24 
"  b^' 

fc2  = 

24 

tan  29° 
log  6  =  i  (log  24  +  colog  tan  29°). 
log  24         =  1.38021 
colog  tan  29°  =  0.25625 


log& 
b 


2)  1.63646 
=  0.81823 
=  6.58. 


TEACHERS     EDITION. 


35 


tan  29°  =  ^. 

0 


a  =  &  tan  29°. 
log  a  =  log  b  4-  log  tan  29°. 

log  6  =  0.81823 

log  tan  29°  =  9.74375 
log  a  =  0.56198 

a  =  3.6474. 

•         A  « 


c  = 


sin  29° 
log  c  =  log  a  +  colog  sin  29°. 

log  a  =  0.56198 

colog  sin  29°  =  0.31443 

log  c  =  0.87641 

c  =  7.5233. 


56.    Given  i^=  100,  c  =  22;  solve 
the  triangle. 

F=  1  ab  =  100. 
ab  =  200. 

2_00 
b  ' 
40000 

62 


a  = 


o3  +  52  =  c2  =  484. 
Substitute, 
40000 


62 


+  62  =  484. 


40000 +  6*  =  484  62. 
6* -484  62  =  -40000. 
¥-{ ) +(242)2=  18564. 


log\/l8564  =  2)4.26867 
=      2.13434 ; 
but    2.13434  =  log  136.25. 
.-.  62  -  242  =  136.25. 
62  =  378.25. 
log  6  =  Hlog  378.25). 
=  1.28889. 
6  =  19.449. 

cos  A  =  -' 
c 

log  COS  A  =  log  6  +  colog  c. 

log  6         =  1.28889 
colog  c         =  8.65758 
log  cos  A  =  9.94647 
A  =  27°  52^ 

B  =62°    8^ 


sin 

A 
a 

—  ^. 

c 

=  c  sin  A. 

log 

5« 

= log  c  + log 

sin 

A 

log 

c 

=  1.34242 

log 

sin 

A 

=  9.66970 

log 

a 

=  1.01212 

a 

=  10.283. 

57.  Find  the  angles  of  a  right 
triangle  if  the  hypotenuse  is  equal 
to  three  times  one  of  the  legs. 

Let  c  =  hypotenuse, 

and  let         c  =  three  times  a,  one  of 
the  legs. 

•        A  O, 

sm  A  =  -' 
c 

log  sin  A  =  log  a  +  colog  c. 


TRIGONOMETRY. 


log  a 

=  0.00000 

colog  c 

=  9.52288 

-10 

log  sin 

A 

=  9.52288 

A 

=  19°  28^ 

17^^. 

B 

=  70°  3F 

43^^ 

58.  Find  the  legs  of  a  right  tri- 
angle if  the  hypotenuse  =  6,  and 
one  angle  is  twice  the  other. 

Let         c  =  hypotenuse  =  6, 
and  let      B  =  twice  A  ; 
then  B  =  60°, 

A  =  30°. 


sin  A 


a  =  c  sin  A. 
log  a  =  log  c  +  log  sin  A. 

log  c         =  0.77815 
log  sin  A  =  9.69897 


log 
a 

a 

=  0.47712 
=  3. 

sin 

B 
h 

h 
~  c 

=  c  sin  B. 

log 

h 

=  log  c  +  log 

sin  B 

log 

c 

=  0.77815 

log 

sin 

B 

=  9.93753 

log 

h 

=  0.71568 

h 

=  5.1961. 

59.   In  a  right  triangle  given  c, 
and  A  =  nB  ;  find  a  and  h. 
B=^90°-A 
=  90°  -  nB. 
B{n  +  1)  =  90°. 
90° 


B  = 


n  +  l 


cos  B 

a 

~  g' 



90° 

a 

cos 

71  +  1 

c 

90° 

a 

=  c  cos 

h 

n  +  l 

sin  5 

c 

.       90° 

_h 

sm 

n  +  l 

c 

b 

90° 

=  c  sir 

L -. 

n  +  l 

60.  In  a  right  triangle  the  differ- 

ence between  the 

hypotenuse 

and 

the  greater  leg  is 

equal  to  the 

jdif- 

ference  between  tt 

e  two  legs ; 

find 

the  angles. 

c  —  a 

=  a  — 

6. 

2a-b 

=  c. 

(1) 

a2  +  62 

=  c2. 

(2) 

Squaring 

(1). 

4o2- 

-4a& 

+  62  =  c2 

+  62  =  c2 


3a2_4a6 

=  0 

3a2  =  4a&. 

3a  =  46. 

n        4& 

tan^  =  -=i 
h     3 

log  tan  ^  =  log  4  +  colog  3. 

log  4        =   0.60206 
colog  3         =    9.52288  -  10 
log  tan  ^  =  10.12494 
A  =53°    7M8/^. 

B  =  36°  52^  12^^ 


TEACHERS     EDITION. 


37 


61.  At  a  horizontal  distance  of 
120  feet  from  the  foot  of  a  steeple, 
the  angle  of  elevation  of  the  top 
was  found  to  be  60°  30^ ;  find  the 
height  of  the  steeple. 

tan  A  =  — 
b 

a  =  b  tan  A. 
log  a  =  log  b  +  log  tan  A. 

log  5  =  2.07918 
log  tan  A  =  10.24736 
log  a  =  2.32654 
a  =  212.1. 

62.  From  the  top  of  a  rock  that 
rises  vertically  326  feet  out  of  the 
water,  the  angle  of  depression  of  a 
boat  was  found  to  be  24° ;  find  the 
distance  of  the  boat  from  the  foot 
of  the  rock. 

cot  A  =  — 
a 

b  =  a  +  cot  A. 

log  b  =  log  a  +  log  cot  a. 

log  a  =  2.51322 
log  cot  A  =  10.35142 
log  6  =  2.86464 
6  =  732.22. 

63.  How  far  is  a  monument,  in 
a  level  plain,  from  the  eye,  if  the 
height  of  the  monument  is  200  feet 
and  the  angle  of  elevation  of  the 
top  3°  30^  ? 

cot  J.  =  — 
a 

b  =  a  cot  A. 
log  6  =  log  a  +  log  cot  A. 


log  a        =  2.30103 
log  cot  J.  =  1.21351 

log  b         =  3.51454 
b  =  3270. 

64.  In  order  to  find  the  breadth 
of  a  river  a  distance  AB  was  meas- 
ured along  the  bank,  the  point  A 
being  directly  opposite  a  tree  C  on 
the  other  side.  The  angle  ABC 
was  also  measured.  If  AB  =  96 
feet,  and  ABO=  21°  14^  find  the 
breadth  of  the  river. 

If  ABC=  45°,  what  would  be  the 
breadth  of  the  river  ? 

tsin  B  =  AC ^AB. 

AC=ABxt&nB. 

log  AC  =  log  AB  +  log  tan  B. 

log  AB     =  1.98227 
log  tan  B  =  9.58944 

log^C     =1.57171 
AC  =  37.3  feet. 


log  AC  =  log  AB  +  log  tan  B. 

log  AB    =    1.98227 
log  tan  B  =  10.00000 

log^C     =    1.98227 
AC  =96  feet. 

65.  Find  the  angle  of  elevation 
of  the  sun  when  a  tower  a  feet  high 
casts  a  horizontal  shadow  b  feet 
long.  Find  the  angle  when  a  =  120, 
5  =  70. 

tan^  =  -. 
b 


tan  A 


120 

70 ' 


log  tan  ^  =  log  120  +  colog  70. 


38 


TRIGONOMETRY. 


log  120     =    2.07918 
colog  70       =    8.15490  -  10 
log  tan  J.  =  10.23408 
A  =  59°  44^  35^^ 

66.  How  high  is  a  tree  that  casts 
a  horizontal  shadow  b  feet  in  length 
when  the  angle  of  elevation  of  the 
sun  is  A°  ?  Find  the  height  of  the 
tree  when  b  =  80°,  A  =  50°. 

tan  A  =  -• 

0 

a  =  h  tan  A. 
log  a        =  log  b  +  log  tan  A. 

log  6  =  1.90309 
log  tan  A  =  10.07619 
log  a  =  1.97928 
a  =  95.34. 

67.  What  is  the  angle  of  eleva- 
tion of  an  inclined  plane  if  it  rises 
1  foot  in  a  horizontal  distance  of  40 
feet? 

tan  A  =  -' 

0 

log  tan  A  =  loga  +  colog  b. 
log  a        =  0.00000 
colog  b         =  8.39794-10 
log  tan  A  =  8.39794 
A  =1°  25'  56'^ 

68.  A  ship  is  sailing  due  north- 
east with  a  velocity  of  10  miles  an 
hour.  Find  the  rate  at  which  she 
is  moving  due  north  and  also  due 
east. 

Let  AB  be  the  direction  of  the 


vessel,  and  equal  one  hour's  progress 
=  10  miles. 

A  C=  distance  due  east  passed  over 
in  one  hour. 

As  the   direction  of  the  ship  is 
north-east, 

A  =  45°. 
b  =  c  cos  A. 
log  b  =  log  c  +  log  cos  A. 

log  10  =  1.00000 
log  cos  A  =  9.84949 
log  b  =  0.84949 
b  =  7.0712  miles  due  east, 

and  also  due  north,  since 
AP=  AO. 


69.   In  front  of  a  window  20  feet 

high  is  a  flower-bed  6  feet  wide. 
How  long  must  a  ladder  be  to  reach 
from  the  edge  of  the  bed  to  the 
window  ? 

tan  J.  =  -• 

0 

log  tan  A  =  log  20  +  colog  6. 

log  20       =   1.30103 
colog  6         =    9.22185-10 
log  tan  A  =  10.52288 
A  =  73°  18^ 

a 
c  = 


sin -4 
log  c  =  log  20  -f-  colog  sin  A. 

log         a  =  1.30103 
colog  sin  A  =  0.01871 
log  c         =  1.31974 
c  =  20.88. 


TEACHERS     EDITION. 


39 


70.  A  ladder  40  feet  long  may  be 
so  placed  that  it  will  reach  a  win- 
dow 33  feet  high  on  one  side  of  the 
street,  and  by  turning  it  over  with- 
out moving  its  foot  it  will  reach  a 
window  21  feet  high  on  the  other 
side.  Find  the  breadth  of  the 
street. 

p     33 

cos  B  =^  — 

40 
log  33       =  1.51851 
colog  40       =  8.39794  -  10 

log  cos  B  =  9.91645 

B  =  34°  2V  45^^ 

tan  5  =  —. 


33 

• 

b  = 

=  33  tan  B 

log 

33 

=  1.51851 

log 

tan^- 

=  9.83571 

log 

b 

- 1.35422 

b  = 

=  22.605. 

cos 

B'= 

=  21. 

40 

log  21  ^    =  1.32222 
colog  40  *    =8.39794-10 
log  cos  B^=  9.72016 
B^  =  58°  19^  54^^ 

tan  B'=  — . 
21 

6^=  21  tan  B^. 

log  21       =  1.32222 
log  tan  ^^=0.20982 

log  ¥       =  1.53204 

b^  =  34.044 

h  =  22.605 

b  +  y  56.649 


71.  From  the  top  of  a  hill  the 
angles  of  depression  of  two  succes- 
sive milestones,  on  a  straight  level 
road  leading  to  the  hill,  are  observed 
to  be  5°  and  15°.  Find  the  height 
of  the  hill. 


5280 


sin  5°  = 


5280 
a  =  5280  sin  5°. 

log  5280  =3.72263 
log  sin  5°  =  8.94030 
log  a         =  2.66293 

sin  10°=  -. 
c 

a  =  c  sin  10°. 

c  =  -^L_. 
sin  10° 

log  a        =  2.66293 

colog  sin  10°=  0.76033 

logc         =  3.42326 

cos  75°=  ^. 
c 

J  =  c  cos  75°. 

log  c         =  3.42326 

log  cos  75°=  9.41300 

log  b         =  2.83626 

b  =  685.9  feet 

-  228.63  yards 

72.  A  fort  stands  on  a  horizontal 
plane.  The  angle  of  elevation  at  a 
certain  point  on  the  plane  is  30°, 


40 


TRIGONOMETRY. 


and  at  a  point  100  feet  nearer  the 
fort  it  is  45°.    How  high  is  the  fort  ? 


A      100  ft.     A'  C 

Let  B  represent  the  fort,  AC  the 
horizontal  plane,  £C  a  ±  from  fort 
to  plane, 

BAC=  angle  made  by  line  from 
eye  of  observer  =  30°. 

BA^Q=  45°  =  angle  of  elevation 
100  feet  nearer. 

From  A^  draw  A^Nl.  to  AB. 
In  rt.  A  AA^N, 
Z.  NAA^=^  30°, 
and  ^  iV^.4^^  =  60°. 

.-.  NA^=  50  feet. 

/.  AN=  \/(100)2-  (50)2 

=  \/7500  =  50\/3 
=  86.602. 

In  rt.  A  BNA' 

^^  =  cot  NBA'=  cot  15°, 
NA' 


and 

log  NA' 
log  cot  15 

log^iV 

BN 
AN 
AB 


B]Sr=  NA'  cot  15°. 


=  1.69897 
=  0.57195 

=  2.27092 

=  186.60 
=    86.60 

=  273.20 


In  rt.  A  ABC, 
ZBAC=  30°, 
and     Z  ABC=  60°. 

.-.  BC=^AB  =  ^x212>.20 

=  136.60  feet. 

73.  From  a  certain  point  on  the 
ground  the  angles  of  elevation  of 
the  belfry  of  a  church  and  of  the  top 
of  a  steeple  were  found  to  be  40° 
and  51°  respectively.  From  a  point 
300  feet  farther  off,  on  a  horizontal 
line,  the  angle  of  elevation  of  the 
top  of  the  steeple  is  found  to  be 
33°  45''.  Find  the  distance  from 
the  belfry  to  the  top  of  the  steeple. 


Draw  DE  ±  to  AB  from  D. 
In  A  BED 

^  =  sin  33°  45^. 
BD 

^Z)  =  300xsin33°45^ 

log  300  =  2.47712 

log  sin  33°  45^=  9.74474 

log  ED  =  2.22186 

Z  EAD  =  180°-  33°  45^  -  (ISO" 
51°)  =  17°  15^ 
In  A  ADE 

ED 


AD 


=  sin  17°  15^. 


TEACHERS     EDITION. 


41 


AD  = 


ED 


sin  17°  15^ 

log  ED  =  2.22186 

colog  sin  17°  15^=  0.52791 

log  AD  =  2.74977 


=  cos  51°. 


In  A  ADC 

DC 

AD 

DC=  AD  cos  51°. 
log  AD  =  2.74977 

log  cos  51°       =  9.79887 
log  DO  =  2.54864 


In  A  ADO 


^=  tan  51°. 
DC 

AC=DOiQ.uf>l°. 


log  DC 

=   2.54864 

log  tan  51° 

=  10.09163 

log^C 

=   2.64027 

AO 

=  436.79 

In  A  EDO 

EC 
DC 

=  tan  40°. 

EC 

=  DC  tan  40 

log  DC 

=  2.54864 

log  tan  40° 

=  9.92381 

log  EC 

=  2.47245 

EC 

=  296.79. 

AC- EC 

=  :4o. 

74.  The  angle  of  elevation  of  the 
top  of  an  inaccessible  fort  C,  ob- 
served from  a  point  A,  is  12°.  At 
a  point  £,  219  feet  from  A  and  on 
a  line  AB  perpendicular  to  AC,  the 
angle  ABC  is  61°  45^.  Find  the 
height  of  the  fort. 


In  rt.  A  CAB 


^  =  cot  ABC 
AC 

AC  = 


AB 


cot  ABC 
log  AC  =  log  AB +  co\ogcot  ABC. 

log  AB  =  2.34044 

colog  cot  ABC    =  0.26977 

log  AC  =  2.61021 

In  rt.  ^ADC 

^  =  sin  CAD. 
AC 

CD  =  AC  sin  CAD. 
log  CD  =  log  AC+  log  sin  CAD. 

log  AC  =2.61021 

log  sin  CAD  =  9.31788 

log  CD  =  1.92809 

CD  =  84.74  feet. 


42 


TRIGONOMETRY. 


Exercise  VII.     Page  25. 


1. 

In 

an  isosceles  triangle, 

given 

a 

and  A 

;  find  C,  c,  h. 

C=  180°  -2  A 

=  2  (90°  -  A). 

i^  =  cos  A. 

a 

c  =  2  a  cos  J.. 

^        •      A 
-  =  sm  -4. 

a 

h  =  a  sin  A. 

2.    In  an  isosceles  triangle,  given 
a  and  C;  find  J.,  c,  A. 


C 

+  2^  = 

180°. 

^  = 

90°- i  (7. 

a 

cos  A. 

c  = 

2  a  cos  A. 

h_ 
a 

sin  A. 

h  = 

a  sin  J.. 

3. 

In 

an  isosceles  triangle, 

c  and  A 

;  find  C,  a,  h. 

C^ 

180°  -2  A 

= 

-2{dO°-A). 

Jc 

^r.c.        A 

4.   In  an  isosceles  triangle,  given 
c  and  C;  find  A,  a,  h. 

^  =  90°  -  J  a 

^^  A 

*-  =  cos  J.. 
a 

r. 

a  = 


2  cos  J. 


given 


2a  = 


a  = 


cos  J. 

^ 

2  COS  J. 


—  =  sm  A. 
a 

h  =  asm  A. 


5.    In  an  isosceles  triangle,  given 
h  and  A ;  find  C,  a,  c. 

(7=2(90°-^). 

sm  A  =  — 
a 

.'.  a  =  h-T-  sin  A. 

a      2a 
.'.  c  =  2a  COS  A, 


6.   In  an  isosceles  triangle,  given 
Ih  and  C\  find  J.,  a,  c. 

J.  =  90°  -  J  G 
sm  -d  =  — 


-  =  sin  A. 
a 

A  =  a  sin  J.. 


a 

a  = 

h  -T-  sin  A. 

cos  ^  = 

2-2  =_^. 
a      2a 

c  = 

2  a  cos  J.. 

7.   In 

a  and  ^ 

an  isosceles  triangle, 
;  find  A,  C,  c. 

given 

sin  J.= 

Ah-  a. 

C  = 

180°  -2  A. 

TEACHERS     EDITION. 


43 


=  2(90°-^). 

a      2a 
c  =  2a  cos  A. 

8.   In  an  isosceles  triangle,  given 
c  and  h ;  find  A,  C,  a. 


tan  A  = 

(7= 

-180°-2^ 

= 

=  2(90°-^). 

sin 

A  = 

h 

a  = 

=  A  -f-  sin  A. 

9.   In 

an 

isosceles  triangle,  given 

a  =  14.3 

,  c 

=  11 

;  find  A,  C,  U. 

cos 

A  = 

a 

log 

COS 

A  = 

--  log  ^c  +  colog  a. 

log 

^c 

= 

-  0.74036 

colog 

a 

= 

=  8.84466  - 10 

log 

cos 

A  = 

=  9.58502 

A 

= 

67°  22^  50^^ 

C 

= 

2  (90°  -  A) 

= 

45°  14^  20^^ 

sin 

A  = 

a 

h  = 

a  sin  A. 

log 

h  = 

log  a  +  log  sin  A. 

log 

a 

= 

1.15534 

log 

sin 

A  = 

9.96524 

log 

h 

= 

1.12058 

h 

^ 

13.2. 

10.  In  an  isosceles  triangle,  given 
a  =  0.295,  A  =  68°  10' ;  find  c,  A, 
F. 


sin 

A  = 

h 

a 

h  = 

=  a  sin  A. 

logA  = 

=  log  a  +  Ic 

)g  sin  A 

log  a 

=  9.46982  - 

-10 

log  sin 

A  = 

=  9.96767 

log  A 

=  9.43749  - 

10 

h 

=  0.27384. 

cos 

A  = 

a 

ic  = 

--  a  cos  A. 

log 

\c  = 

=  log  a  +  Ic 

>g  cos  A 

log  a 

=  9.46982  - 

10 

log  cos 

A  = 

=  9.57044 

logic 

=  9.04026  - 

10 

ic 

=  0.109713. 

c 

=  0.21943. 

F= 

^ich. 

2F^ 

■-  ch. 

\q^2F= 

■■  log  c  +  lo 

g^. 

logc 

-9.34130- 

10 

log  A 

=  9.43749  - 

10 

Iog2i^ 

r 

8.77879  - 

10 

2F 

= 

0.060089. 

F 

0.03004. 

11.  In  an  isosceles  triangle,  given 
c  =  2.352,  (7=69°49^  find  a,  h, 
F. 

I  C=  34°  54'  30''. 

siii|C=l^. 

a 


a  = 


+  c 


sm  I  C 
loga  =  log|c+colog  sin  |C. 


44 


TRIGONOMETKY. 


log  Jc       = 

-  0.07041 

cologsin  J(7= 

=  0.24240 

log  a 

=  0.31281 

a              = 

=  2.0555. 

cos  1 C^ 

h 

a 

h  = 

=  a  cos  ^  0. 

log  A  = 

log  a  +  log  cos 

io. 

log  a 

=  0.31281 

log  cos  ^  C= 

=  9.91385 

log  h        =  0.22666 
h  =  1.6852. 

2  F=  ch. 
log  2  i^=  log  c  +  log  A. 

log  c  =  0.37144 

log  h  =  0.22666 

log2i?'  =0.59810 

2F  =  3.9637. 

jP  =  1.9819. 


12.  In  an  isosceles  triangle,  given 
A  =7.4847,  ^=76°  14^;  find  a,  c, 
F. 

sin  A  =  -' 


sin -4 

log  a 

=  log  h  +  colog  sin 

A. 

log 

h 

=  0.87417 

colog 

sin  A 

=  0.01266 

log 

a 

=  0.88683 

a 

=  7.706. 

tan  A=—' 
Jc  = 


tan  .4 
log  J  c  =  log  A  +  colog  tan  A 

logh        =0.87417 
colog  tan  A  =  9.38918  -  10 

log^c       =0.26335 
^c  =1.8338. 

c  =  3.6676. 

F=^ch. 
log  i^=log|c  +  log/i. 

logjc  =0.26335 

log  h  =  0.87417 

log  i^  =  1.13752 

F  =  13.725. 

13.  In  an  isosceles  triangle,  given 
a  =  6.71,  A  =  6.60;  find  ^,  C,  c. 

8inJ.  =  ^. 
a 

log  sin  A  =  log  h  +  colog  a. 

log  /i        =  0.81954 
colog  a         =9.17328-10 

log  sin  A  =  9.99282 

A  =  79°  36^  30^^ 

0  =  20°  47^ 

C0S7l  =  4^. 

a 

^c  =  acosA. 
log  J  c  =  log  a  +  log  cos  A 
log  a  =  0.82672 
log  cos  A  =  9.25617 
log^c  =0.08289 
|c  =1.2103. 

c  =  2.4206. 


TEACHERS     EDITION. 


45 


14.  In  an  isosceles  triangle,  given 
c  =  9,  ^  =  20 ;  find  A,  c,  a. 


tan 


^C=i^. 


log  tan  ^  c  =  log  J  c  +  colog  A. 

log^c      =0.65321 
colog  h         =  8.69897  -  10 


log 

tan  ^(7= 

--  9.35218 

hG 

=  12°  40^  49^^ 

C 

=  25°  2V  38^^ 

2A  = 

=  180°  -  a 

A^ 

-  77°  19^  11^^ 

sin  ^  = 

h 
a 

a  = 

h 

sin  JL 
log  a  =  log  A  +  colog  sin  A. 

log  h        =  1.30103 
colog  sin  A  =  0.01072 

log  a        =  1.31175 
a  =  20.5. 

15.  In  an  isosceles  triangle,  given 
c  =  147,  i^=  2572.5;  find^.C^^- 

c 

log  h  =  log  2  F+  colog  c. 

log2i?'     =3.71139 
colog  c         =  7.83268-10 

log^         =  1.54407 
h  =35. 

tan  J.  =  — 

log  tan  A  =  log  h  +  colog  ^  c. 


log 
colog 
log 
A 
G 


h        =  1.54407 

|c       =  8.13371  -  10 

tan  A  =  9.67778 

=  25°  28^ 

=  2  {90° -A) 

=  129°  4^ 


a  = 


h 


log 

log 
colog 
log 
a 


sin.  A 
=  log  ^  +  colog  sin  A. 

=  1.54407 


sin  A  =  0.36655 
a        =  1.91062 
=  81.41. 


16.   In  an  isosceles  triangle,  given 
h  =  16.8,  F=  43.68  ;  find  A,  C,  a,  c. 


ic 

F 

h 

log^c 

=  log  F  -\-  colog  h 

logi^ 

=  1.64028 

colog  h 

=  8.77469-10 

log^c 

=  0.41497 

^G 

=  2.60. 

G 

=  5.2. 

tan  A  = 


A 


log  tan  A  =  log  h  +  colog  ^  c. 

log^        =   1.22531 
colog  I  c       =    9.58503  - 10 
log  tan  A  =  10.81034 
A  =  81°  12^  9^^ 

1  C  =8°  47^  51^^ 

C  =  17°  35^  42^^ 


46 


TRIGONOMETRY. 


COS  A  = 


^c 


log  a        =  log  I  c  +  colog  cos  A. 

logic      =0.41497 
colog  cos  A  =  0.81547 
log  a        =  1.23044 
a  =17. 

17.    In  an  isosceles  triangle,  find 
the  value  of  F  in  terms  of  a  and  c. 


F=  J  ch. 


=v 


4  a2  -  c2 


=  *V4a2-c2. 


■Z^=2c(jV4a2-c2) 


=  |c\/4a2-c2. 

18.  In  an  isosceles  triangle,  find 
the  value  of  F  in  terms  of  a  and  C. 

F=  I  ch. 

J  c  =  a  sin  f  C. 

h  =  a  cos  J  0. 

F=  a  sin  ^  Cx  a  cos  J  C. 

=  a^  sin  J  C  cos  J  C. 

19.  In  an  isosceles  triangle,  find 
the  value  of  F  in  terms  of  a  and  A. 

F=lch. 
ic=  a  cos  A. 
h  =  a  sin  A. 
F=  a  cos  Ax  a  sin  A 
=  a^  sin  J.  cos  A. 

20.  In  an  isosceles  triangle,  find 
the  value  of  F  in  terms  of  h  and  C. 


F=  i  cA. 
J  c  =  A  tan  J  C. 

J'=A(/itaniC) 
=  h^  tan  J  C. 

21.  A  barn  is  40  X  80  feet,  the 
pitch  of  the  roof  is  45°;  find  the 
length  of  the  rafters  and  the  area 
of  both  sides  of  the  roof. 

40-2  =  20  =  ^c. 

cos^  =  fc^a 

=  20  -f-  a. 

20  =  a  cos  A. 

20 
a  = 

cos -4 

log  a  =  log  20  +  colog  cos  A. 

log  20       =  1.30103 
colog  cos  A  =  0.15051 

log  a         =  1.45154 

a  =  28.284. 

28.284  X  80  =  2262.72. 
2262.72  X  2  =  4525.44. 

22.  In  a  unit  circle  what  is  the 
length  of  the  chord  corresponding 
to  the  angle  45°  at  the  centre  ? 

sin  J  (7=  i^. 
a 

log  J  c  =  log  a  -j-  log  sin  J  C. 

log  a        =  0.00000 

logsin  ^(7=  9.58284 

logjc      =9.58284-10 

Jc  =0.382683. 

G  =  0.76537. 

23.  If  the  radius  of  a  circle  =  30, 
and  the  length  of  a  chord  =  44,  find 
the  angle  at  the  centre. 


TEACHERS     EDITION. 


47 


sinK'=— • 
a 

logsin  JC=  log  2^  c  +  colog  a. 
logic       =1.34242 
colog  a         =  8.52288  -  10 
log  sin  ^(7=  9.86530 
IC  =47°  10^ 


C 


=  94°  20^ 


24.  Find  the  radius  of  a  circle  if 
a  chord  whose  length  is  5  subtends 
at  the  centre  an  angle  of  133°. 

sin  i  (7=  i^- 
a 

log  a  =  log  J  c  +  colog  sin  J  C. 

logic       =0.39794 

colog  sin -^(7=0.03760 

log  a         =0.43554 

a  =  2.7261. 


25.  What  is  the  angle  at  the  cen- 
tre of  a  circle  if  the  corresponding 
chord  is  equal  to  |  of  the  radius  ? 


Let  a  =  3,  then  c  =  2,  and  J  c  =  1. 

sin  h  C=  — 
^         3 

log  sin  i  C=  log  1  +  colog  3. 

log  1  =  0.00000 

colog  3  =  9.52288  -  10 


log  sin  J  (7=  9.52288 
i  C  =19°  28^ 

C  =  38°  56^  33^^^ 


19°  28^  16|^^. 


26.  Find  the  area  of  a  circular 
sector  if  the  radius  of  the  circle  =  12 
and  the  angle  of  the  sector  =  30°. 

Area  O  =  irE^. 


Area  sector  = 


360 


log  area  sector  =  log  30  -f  colog 
360  +  log  TT  +  2  log  R. 


log  30 

=  1.47712 

colog  360 

=  7.44370  - 

-10. 

logTT 

=  0.49715 

2  log  B 

=  2.15836 

log  area 

=  1.57633 

Area 

=  37.699. 

ExEECisE  VIII.     Page  26. 


1.    In   a  regular  polygon  given 
n  =  10,  c  =  1  ;  find  r,  A,  F. 


i(7= 


180= 


=  18°. 


10 
i  e  =  0.5. 
A  =  72°. 
A  =  2  c  tan  A. 
log  h  =  log  i  c  +  log  tan  A. 


logic      =    9.69897-10 
log  tan  A  =  10.48822 

log  h        =    0.18719 
h  =   1.5388. 

log  r  =  log  i  c  +  colog  cos  A. 
logic      =9.69897-10 
colog  cos  A  =  0.51002 

log  r         =  0.20899 
r  =  1.618. 


48 


TE-IGONOMETRY. 


F 

=  hhp. 

log  h 

=  0.18719 

\ogp 

=  1.00000 

\og2F 

=  1.18719 

2F 

=  15.388 

F 

=  7.694. 

2.   In 

a 

regular  polygon  given 

n=l2,p 

r 

10  ■  find  r,  h,  F. 

'^G 

=  15°. 

A 

=  75°. 

c 

=  70  H- 12  =  5.833. 

\c 

=  2.917. 

h 

=  J  c  tan  A. 

logic 

=    0.46494 

log  tan  A 

=  10.57195 

log  A 

=    1.03689 

h 

=  10.886. 

r 

=  1  c  cos  A. 

log  |c 

=  0.46494 

Colog  COS 

A 

=  0.58700 

log  r 

=  1.05194 

r 

=  11.27. 

F 

=  lhp. 

log  A 

=  1.03689 

logp 

=  1.84510 

A  =  r  sin  A. 
log  r  =  0.00000 
log  sin  A  =  9.99335 

log  h         =  9.99335 
h  =  0.9848. 


^  c  =  r  cos  A. 

log  r         =  0.00000 

log  cos  A  =  9.23967 

log  i  c      =  9.23967  - 

-10 

Jc             =0.17365. 

p               =  6.2514. 

F=^hp. 

log  h         =  9.99335  - 

-10 

logp         =0.79598 

log  2  i?'     =  0.78933 

2F           =6.1564. 

F             =  3.0782. 

log  2  i?'     =  2.88199 
2F  =  762.07. 

F  =  381.04. 

3.   In   a  regular  polygon  given 
n  =  18,  r  =  1  ;  find  h,  p,  F. 

^C=  10°. 
A  =  80°. 


4.   In  a  regular  polygon  given 
n  =  20,  r  =  20 ;  find  h,  c,  F. 

iC=    9°. 
A  =  81°. 
7i  =  r  sin  A. 
log  r         =  1.30103 
log  sin  A  =  9.99462 

log  h        =  1.29565 
h  =  19.754. 

^  c  =  r  cos  -4. 
log  r  =  1.30103 
log  cos  A  =  9.19433 
log  J  c  =  0.49536 
^c  =3.1286. 

c  =  6.257. 

p  =  125.14. 


TEACHERS     EDITION. 


49 


F^lhp. 

log  A 

=  1.29565 

logp 

==  2.09740 

\og2F 

=  3.39305 

2F 

=  2472. 

F 

=  1236. 

tan 


1  /nr_  2  <^ 

2  O  —  — - 


5.  In  a  regular  polygon,  given 
n  =  8,  h  =  l]  find  r,  c,  F. 

^C=22°30^ 
tan^a=t^. 

log  J c  =  log  ^+log  tan  \C. 

log  h  =  0.00000 

log  tan  ^  C  =  9.61722 

log^c  =9.61722-10 

^c  =  0.41421. 
c  =  0.82842. 

cos  I  (7=-. 
r 

log  r  =  log  ^.  +  colog  cos  \  C. 

log  h  =  0.00000 

colog  cos  ^  C  =  0.03438 

log  r  =  0.03438 

r  =  1.0824. 

F=ihp. 
=  3.3137. 

6.  In  a  regular  polygon,  given 
n  =  11,  F=  20  ;  find  r,  /i,  c. 

^0=ph. 


^  =  S' 


A  =  40. 


|a=16°22^ 


h 


Substituting  values  of  h  and  c, 

tania=^^^  =  ^. 
^         22      p      880 

logi>  =  i  (log  880  +  log  tan  J  C). 

log  880         =     2.94448 

log  tan  ^  a  =     9.46788 

2)  2.41236 
logp  =     1.20618 

p  =       16.076. 

c  =         1.4615. 


sin  2  0= 

r 

log 

r  =  log  \c-\-  colog  sir 

iC 

log 

|c 

:  9.86376  - 

10 

colog 

sin  1^  C   = 

=  0.55008 

log 

r            = 

-  0.41384 

r 

= 

=  2.592. 

cos  1  (7= 

r 

log  A  =  log  r + log  cos  \  0. 

log  r  =  0.41384 

log  cos  ^  C   =  9.98204 

log  h  =  0.39588 

h  =  2.4882. 

7.  In  a  regular  polygon,  given 
n  =  7,  i^=  7  ;  find  r,  ^,  p. 

14  =pA. 

P 

|C=25°43^ 


50 


TRIGONOMETRY. 


tan  J  (7=  t^. 
h 


14 


tanJ(7=^-^-L^=^. 
14      p      196 

logp  =  I  (log  196  +  log  tan  ^  C). 

log  196         =  2.29226 
log  tan  ^C  =9.68271 


2)  1.97497 

\ogp 

=  0.98749 

p 

=  9.716, 

^c 

=  0.694. 

tan  J 

C=i-'. 

log  h  =  log  I  c  +  colog  tan  |  C. 

log  f  c  =  9.84136  -  10 

colog  tan  ^C  =0.31729 


\ogh 

=  0.15865 

h 

=  1.441. 

siniC=i^- 
r 

logr  = 

log  J  c  +  colog  sin  1  C. 

log  Jc 

=  9.84136  - 

10 

colog  sin 

|(7  =0.36259 

log  r 

=  0.20395 

r 

=  1.5994. 

8.   Find   the  side   of    a  regular 
decagon  inscribed  in  a  nnit  circle. 

sin  J  a=  i^. 
r 

log  c  =  log  2+log  sin  J  C. 

log  2  =  0.30103 

log  sin  J  (7  =  9.48998 

log  c  =  9.79101  -  10 

c        .  =0.6181. 


9. 

decag 
circle 


log 
log 

log 
ic 
c 


Find   the  side   of   a  regular 
;on  circumscribed  about  a  unit 

^C=18°. 

taniC=i^. 
h 

log  J  c  =  log  h  +  log  tan  |  C. 

h  =0.00000 

tan  J  C  =  9.51178 

^c  =9.51178-10 

==  0.32492. 
=  0.64984.  ^ 


10.  If  the  sides  of  an  inscribed 
regular  hexagon  is  equal  to  1,  find 
the  side  of  an  inscribed  regular  do- 
decagon. 

Let  0  be  the  centre  of  the  circle, 
BC  a  side  of  the  hexagon,  and  BA 
a  side  of  the  dodecagon.  Also  let 
OD  be  ±  to  BA. 

Then      OB  =  5C=  1.'' 

Z  BOD  =  15°.  : 

In  rt.  A  ODB 

sin  BOD  =  ^AB^  OB 

AB  ^  2  OB  X  sin  BOD. 

log  AB  =  log  2  0B  +  log  sin  BOD. 

log  2  05  =  0.30103 

log  sin  15°  =  9.41300 

log  AB  =  9.71403  -  10 

AB  =0.51764.-      - 

11.  Given  n  and  c,   and  let  b 

denote  the  side  of  the  inscribed  reg- 
ular polygon  having.  2  w  sides  ;  find 
b  in  terms  of  n  and  c. 

Let  O  be  the  centre  of  the  cir- 
cle,  BG  ih^  j3i4e  of  the  polygon 


TEACHERS     EDITION. 


51 


having  n  sides,  BA  the  side  of  the 
polygon  having  2n  sides.  Then 
OA  is  ±  to  5  C  at  its  middle  point 
D. 

360°      180° 


ZBOA^ 


ZOBC=dO°- 


2n         n 
180° 


n 


The  A  BOA  is  isosceles. 
.\/.OBA=-(l^Q)' 

Q0° 


180^ 
n 


A  ABC=  Z  OB  A-  Z  OBC 
180 


/9Qo_90^\     ^90o_ 


n 


^90° 
n 

ic  90° 

i—  =  cos " 

b  n 

Whence, 


1        J.       90° 
^c  =  0  cos 

n 


h  = 


he 


cos 


90° 
n 


o       90° 
2  cos  — 


n 


12.  Compute  the  difference  be- 
tween the  areas  of  a  regular  octa- 
gon and  a  regular  nonagon  if  the 
perimeter  of  each  is  16. 

^        2n     16 

1  QAO 

^  =  i^=22°30^ 
n 

,  log  A  =  log  I  c  +  log  cot  A. 

logjc  =   0.00000 

log  cot  ^     =10.38278 

log  A  =    0.38278 


log  F=  log  h  +  log  Ijp. 


log  h 
logi^ 


=  0.38278 
=  0.90309 


Ic 


=  1.28587 
=  19.3139.   • 

^=-^  =  —  =  0.8889. 
2n^     18 


A'= 


180° 


20°. 


n' 


log  A^=  log  ^  c'-f  log  cot  A\ 


logic^ 

=   9.94885-10 

log  cot  A' 

'    =10.43893 

logA^ 

=    0.38778 

log 

F^=  log  h^+  log  Ip. 

log  A^ 

=  0.38778 

log^i5 

=  0.90309 

logi^-' 

=  1.29087 

F' 

=  19.5377. 

F'-F 

=  19.5377  - 19.3139 

=  0.2238. 

13.  Compute  the  difference  be- 
tween, the  perimeters  of  a  regular 
pentagon  and  a  regular  hexagon  if 
the  area  of  each  is  12. 


F= 

=  12, 

n  = 

5. 

ic== 

180° 
5 

=  36 

>°. 

F= 

--^hp. 

h  = 

.24 
P 

ic  = 

-.2L- 
2n 

10 

tan 

iC= 

h 

P 
10 

24" 

P' 
240 

p 


52 


TKIGONOMETRY. 


|j2  =  240  tan  J  C. 

log  240        =  2.38021 
log  tan  ^C  =9.86126 


2)2.24147 

logp 

=  1.12074 

p 

=  13.205. 

n  =  6,     I  C'=  30° 

tan 

^^  ~  24"  288 

V' 

p^2=  288  tan  \  C. 

log  288 

=  2.45939 

log  tan 

J  C''=  9.76144 

2)2.22083 

logp^  =1.11042 

f  =12.895. 

p-jt)/  =0.310. 

14.  From  a  square  whose  side  is 
equal  to  1  the  corners  are  cut  away 
so  that  a  regular  octagon  is  left. 
Find  the  area  of  this  octagon. 

1  /360' 


\C^ 


22°  30^ 


A  =  90°-  22°  30' 

=  67°  30^ 


tan  A  = 


\c 


\c^ 


h 


tan  A 
log  I  c  =  log  ^  +  colog  tan  A. 

log  n  =  9.69897  - 10 

colog  tan  ^     =9.61722-10 

logic  =9.31619-10 


log  F=  log  I  c  +  log  ^n. 

logic  =9.31619-10 

logjn  =0.60206 

log  F  =  9.91825  -  10 

F  =  0.82842. 

15.  Find  the  area  of  a  regular 
pentagon  if  its  diagonals  are  each 
equal  to  12. 

C 


ZAOD 

AAOC- 
AACB: 

ZOAD. 
ZDAC= 


180= 


=  36« 


:  180°- 36°  =  144°. 
I  (180° -144°) 

18°  =zaAo. 

90°-ZAOD=5i°. 

54° +18°  =  72°. 


cos  i)^C=^^  = 


AD^ic 
AO     12 


logic 

log  cos  72°  ■■ 
log  12 

logic 


log  12  + log  cos  72°. 

9.48998 
1.07918 

0.56916 


TEACHERS     EDITION. 


tani)^0=— • 
he 

log  h  =  log  J  c  +logtan  54°. 

logjc  =    0.56916 

log  tan  54°  =  10.13874 

log  A  =    0.70790 

p  =  i  c  X  2  n. 

log  F=  log  jc+log  n+log  h. 

logjc  =0.56916 

log  n  =  0.69897 

log  h  =  0.70790 

logi?'  =1.97603 

F  =  94.63. 

16.  The  area  of  an  inscribed  reg- 
ular pentagon  is  331.8;  find  the 
area  of  a  regular  polygon  of  11  sides 
inscribed  in  the  same  circle. 

Let  AB  be  a  side  of  a  regular  in- 
scribed pentagon,  and  AD  the  side 
of  a  regular  inscribed  polygon  of  11 
sides. 

Let  a  be  the  radius  of  the  circle 
whose  centre  is  0,  and  h  and  h^  the 
apothems  of  the  2  polygons,  respec- 
tively. 

Given  F  the  area  of  pentagon 
=  331.8.  Find  F\  the  area  of  11- 
sided  polygon. 

Let  p  and  p^  and  c  and  c'  repre- 
sent the  perimeters  and  sides  of  the 
pentagon  and  11-sided  polygon,  re- 
spectively. 

F==  Iph. 

331.8  =  hph, 
ph  =  663.6. 
663.6 


h 


P 


c  = 

-2. 
5 

ic  = 

10 

ZAOF= 

=  36°. 

tan  36°  =  ^-  =  -2.  X  -2~ 
h      10     663.6 

=    Pi-. 
6636 

log  p2  =  log  tan  36° -h  log  6636. 
log  6636       =  3.82191 
log  tan  36°  =  9.86126 
log  ^2  =3.68317 

logp  =1.84159 

Since       ^  c  =  y^^  of  p, 
log  ^c  =  0.84159. 

sinZ^O^=i^- 
R 

log  R  =  log  J  c  -f  colog  sin  36°. 

log^c  =0.84159 

colog  sin  36°    =0.23078 

log  R  =  1.07237 

Z  A00=  ^^  =  16°21/49^A 

22 


sin  Z  AOC= 

-^c'-^R. 

logR 

=  1.07237 

log  sin  u4  0(7-= 

=  9.44985 

log^c^ 

=  0.52222 

tan^OC= 

Jc^ 

log  h^=  log  I  c^+  colog  tan  AOC. 
logjc^         =0.52222 
colog  tan  ^0C=  0.53220 
log^i^  =1.05442 


54 


TEIGONOMETRY. 


F=  Ip'h' 

=  lc'XllXh' 

log  Jc^ 

=  0.52222 

log  11 

=  1.04139 

log  y 

=  1.05442 

logF 

=  2.61803 

F 

=  414.99. 

17.  The  perimeter  of  an  equilat- 
eral triangle  is  20 ;  find  the  area  of 
the  inscribed  circle. 


Perimeter  ■■ 
AB 
ZOAB 

tan  30°  = 


20. 

ix20  =  -i/. 

i  of  180°  =  30°. 

r 
AB 


log  AB 
log  tan  30° 
log  r 

Area 

log  IT 

log  r^ 
log  area 
Area 


=  0.52288 
=  9.76144 
=  0.28432 

=  irr^, 

=  0.49715 
=  0.56864 
=  1.06579 
=  11.636. 


18.  The  area  of  a  regular  poly- 
gon of  16  sides,  inscribed  in  a  circle, 
is  100 ;  find  the  area  of  a  regular 
polygon  of  15  sides,  inscribed  in  the 
same  circle. 

1(7=360^^110  15/. 
32 

360° 

ia  =  ii^  =  12°. 
^  30 

Let        AC=  h, 

AB  =  r, 

BC^ic  -  _      - 

F=}hp. 

100  =  lhp. 

A  =  20p. 

P 

P. 

32 

200 

l>2  =  6400  tan  J  0. 

log  6400       =  3.80618 
log  tan  I  (7  =  9.29866 


tan  J  C=  rrrx- 


logp 

P 


2)3.10484 
=  1.55242 
=  35.68. 
^0  =  35.68-32 
=  1.115. 


sinJC=^  = 


1.115 


log  1.115     =  0.04727 

cologsiniC   =  0.70976 

log  r  =  0.75703 

—  =  cos  J  (7'' (12°). 
h^  =  r  X  cos  ^  C. 


TEACHERS     EDITION. 


55 


log 
log 

log 


log 
log 

log 


r  =  0.75703 

cos  J  C^  -  9.99040 


ic^ 


=  0.74743 


log 

log 
log 
log 

F 


— -  =  sm  2 
r 

J  c^  =  r  X  sin  J  C\ 

r  =0.75703 

sin  ia  =  9.31788 
^c^         =0.07491 

F=  log  J  c^+  log  n  +  log  h^. 

=  0.07491 

=  1.17609 

=  0.74743 


15 


1.99843 
=  99.640. 


19.  A  regular  dodecagon  is  cir- 
cumscribed about  a  circle,  the  cir- 
cumference of  which  is  equal  to  1 ; 
find  the  perimeter  of  the  dodecagon. 

Given  circumference  of  inscribed 
0  =  1,  n=12;  find  p. 

2  Trr  =  circumference. 


r  = 


circ. 


27r 

\G 

-  3B0°  _  15 
24 

o 

tan  15° 

r 

tan  15° 

3.1416 

log 

tan  15° 

=  9.42805 

colog 

3.1416 

=  9.50284  - 

10 

log 

c 

=  8.93089  - 

10 

log 

12 

=  1.07918 

logp 

=  0.01007 

p 

=  1.0235. 

20.  The  area  of  a  regular  poly- 
gon of  25  sides  is  equal  to  40  ;  find 
the  area  of  the  ring  comprised  be- 
tween the  circumferences  of  the  in- 
scribed and  the  circumscribed  circles. 

ic/i  =  — =  1.6. 
^  25 


^  C=  7°  12^ 

360 


A== 


2n 


=  7°  12^. 


or 


2^  =  tan  I  C, 
h 

i^  =  tan  i  a 

A^  =  -^^. 
tan  2^  C 


log  1.6 
colog  tan  ^  C 

log  h? 
log  h 

h 


=  0.20412 
=  0.89850 

=  1.10262 
=  0.55131. 


r 

r  = 


log  A 

colog  COS  ^  C 

log  r 
log  r^ 


=  COS  ^  07 

h 

cos  I  0 
=  0.55131 

=  0.00344 

=  0.55475 
=  1.10950. 


irr^  =  area  of  circumscribed  O. 

log  TT  =  0.49715 

log  r2  =  1.10950 

log  F  =  1.60665 

F  =  40.425. 

log  TT  =  0.49715 

log  A2  =  1.10262 

logirh^         =1.59977 

Area  =  39.790,(inscrib'dO) 

40.425-39.790  =  0.635. 


56 


TRIGONOMETRY. 


Exercise  IX.     Page  36. 


1.  Construct  the  functions  of  an 
angle  in  Quadrant  II.  What  are 
their  signs  ? 

Sines  and  tangents  extending  up- 
wards from  horizontal  diameter  are 
positive ;  downwards,  negative.  Co- 
sines and  cotangents  extending  from 
vertical  diameter  towards  the  right 
are  positive ;  towards  the  left,  nega- 
tive. Signs  of  secant  and  cosecant 
are  made  to  agree  with  cosine  and 
sine,  respectively.     Hence, 

sin  and  esc  are  -f 
cos  and  sec  are  — 
tan  and  cot  are  — 

2.  Construct  the  functions  of  an 
angle  in  Quadrant  III.  What  are 
their  signs  ? 

sin  and  esc  are  — 
cos  and  sec  are  — 
tan  and  cot  are  -h 

3.  Construct  the  functions  of  an 
angle  in  Quadrant  IV.  What  are 
their  signs  ? 

sin  and  esc  are  — 
cos  and  sec  are  + 
tan  and  cot  are  — 

4.  What  are  the  signs  of  the 
functions  of  the  following  angles  : 
340°,  239°,  145°,  400°,  700°,  1200°, 
3800°  ? 

340°  is  in  Quadrant  IV. 

sin  =  —        tan  =  —        sec  =  + 

cos  =  +        cot  =  —        esc  =  — 


239°  is  in  Quadrant  III. 

sin  =  —        tan  =  +         sec  =  — 
cos  =  —        cot  =  -f         esc  =  — 

145°  is  in  Quadrant  II. 
sin  =  -t-         tan  =  —        sec  =  — 

cos  =  —  cot  =  —  CSC  =  -f- 

400  =  360°  -f  40°  =  signs  of  func- 
tions of  40°. 

40°  is  in  Quadrant  I. 

sin  =  +         tan  =  -f         sec  =  + 

cos  =  +  cot  =  +  CSC  =  -t- 

700°  =  360°  -f  340°  =  signs  of  the 
functions  of  340°. 

340°  is  in  Quadrant  IV. 

sin  =  —        tan  =  —        sec  =  + 
cos  =  +        cot  =  —        esc  =  — 

1200°  =  3  X  360°  +  120°  =  signs 
of  the  functions  of  120°. 

120^  is  in  Quadrant  II. 
sin  =  -h         tan  =  —         sec  =  — 

cos  --=  —  cot  =  —  CSC  =  + 


3800°  =  10  X  360°  +  200° 
of  the  functions  of  200°. 


signs 


200°  is  in  Quadrant  III. 

sin  =  —        tan  =  -f        sec  =  — 

cos  =  —  cot  =  +  CSC  =  — 

5.  How  many  angles  less  than 
360°  have  the  value  of  the  sine 
equal  to  -I-  f,  and  in  what  quad- 
rants'do  they  lie?   . 


TEACHERS     EDITION. 


57 


Since  the  sine  is  +,  by  §  24,  the 
angles  can  lie  in  but  two  quadrants, 
the  first  and  second. 

In  the  first  quadrant,  by  |  3,  the 
sine  increases  from  0  to  1,  and  in 
the  second,  decreases  from  1  to  0. 
This  is  a  continually  increasing  and 
decreasing  quantity. 

Therefore  there  can  be  but  one 
angle  whose  sine  is  equal  to  +  f  in 
each  quadrant,  the  first  and  second. 

6.  How  many  values  less  than 
720°  can  the  angle  x  have  if  cos  x 
=  +f,  and  in  what  quadrants  do 
they  lie  ? 

720°  is  twice  360°;  hence  the 
moving  radius  will  make  exactly  2 
complete  revolutions. 

The  cosine  has  the  +  sign  in  the 
first  and  fourth  quadrants,  hence  it 
will  have  four  values  :  two  in  Quad- 
rant I.  and  two  in  Quadrant  IV. 

7.  If  we  take  into  account  only 
angles  less  than  180°,  how  many 
values  can  x  have  if  sin  a;  =  f  ?  if 
cos  a;  =  ^  ?  if  cos  x  =  —  |  ?  if  tan  x 
=  f?  ifcotaj  =  -7? 

(i.)  Sign  being  +,  the  angle  can 
be  in  Quadrant  I.  or  II. 

.-.  two  values,  one  in  Quadrant  I. 
and  one  in  Quadrant  II. 

(ii.)  Sign  being  -f,  the  angle  is  in 
Quadrant  I.  or  IV. 

.•.  two  values,  one  in  Quadrant  I. 
and  one  in  Quadrant  IV. 

(iii.)  Sign  being  — ,  the  angle  can 
be  in  Quadrant  II.  or  III. 

.•.  two  values,  one  in  Quadrant  II. 
and  one  in  Quadrant  III. 

(iv.)  Sign  being  +,  the  angle  can 
be  in  Quadrant  I.  or  III. 


.•.  two  values,  one  in  Quadrant  I. 
and  one  in  Quadrant  III. 

(v.)  Sign  being  — ,  the  angle  can 
be  in  Quadrant  II.  or  IV. 

.•.  two  values,  one  in  Quadrant  II. 
and  one  in  Quadrant  IV. 

8.  "Within  what  limits  must  the 
angle  x  lie  if  cosx  =  — |?  if  cot  a; 
=  4?  if  sec  a;  =  80?  ifcsca;  =  -3? 
{x  to  be  less  than  360°.) 

If  cos  X  =  —  f ,  X  must  lie  in  the 
second  or  third  quadrant,  or  between 
90°  and  270°. 

If  cot  a;  =  4,  X  is  between  0°  and 
90°,  or  180°  and  270°. 

If  sec  a;  =  80,  a;  is  between  0°  and 
90°  or  270°  and  360°. 

If  CSC  a;  =  —  3,  a;  is  between  180° 
and  360°. 

9.  In  what  quadrant  does  an 
angle  lie  if  sine  and  cosine  are  both 
negative  ?  if  cosine  and  tangent  are 
both  negative  ?  if  the  cotangent  is 
positive  and  the  sine  negative? 

(i.)  Sine  is  negative  in  Quadrants 
II,  and  III.  ;  cosine  is  negative  in 
Quadrants  III.  and  IV. 

.•.  angles  having  both  sine  and 
cosine  negative  are  in  Quadrant 
III. 

(ii.)  Cosine  is  negative  in  Quad- 
rants II.  and  III.  ;  tangent  is  nega- 
tive in  quadrants  II.  and  IV. 

.•.  angles  having  both  cosine  and 
tangent  negative  are  in  Quadrant 
II. 

(iii.)  Cotangent  is  positive  in 
Quadrants  I.  and  III. ;  sine  is  neg- 
ative in  Quadrants  III.  and  IV. 

.'.  angles  having  cotangent  posi- 
tive and  sine  negative  are  in  Quad- 
rant III. 


58 


TRIGONOMETRY. 


10.  Between  0°  and  3600°  how 
many  angles  are  there  whose  sines 
have  the  absolute  value  f  ?  Of 
these  sines  how  many  are  positive 
and  how  many  negative  ? 

Between  0°  and  3600°  there  are 
10  revolutions,  and  in  each  there 
are  4  angles  whose  sines  have  the 
absolute  value  f.  .*.  there  are  40 
angles.  The  sine  is  positive  in 
Quadrants  I.  and  II.,  and  negative 
ing  Quadrants  III.  and  IV.  .•.  there 
are  20  angles  with  the  sine  positive, 
and  20  with  the  sine  negative. 

11.  In  finding  cos  x  by  means  of 
the  equation  cos  x  =  ±  Vl  —  sin^  x, 
when  must  we  choose  the  positive 
sign  and  when  the  negative  sign  ? 

Since  cosines  only  of  angles  in 
Quadrants  I.  or  IV.  are  positive,  we 
use  the  sign  +  only  when  angle  x 
lies  within  these  limits. 

Also,  since  cosines  of  angles  in 
Quadrants  II.  and  III.  are  nega- 
tive, we  use  the  sign  — ,  when  x  is 
known  to  lie  in  either  of  these. 

12.  Given  cos  a;  =  -  V| ;  find  the 
other  functions  when  x  is  an  angle 
in  Quadrant  II. 

sin^x  +  cos^a;  =  1. 


sin  X  =  Vl  —  cos^o; 

=  Vl  -  {-VlY  =  VJ. 

11/9 

esc  X  = =  — r:  =  "v^- 

sm  x      V^ 


sec  x  = 


cos  03 


tan  05  = 


cos  a;      —VI 


=  -V2. 


sin^^_Vj_  ^  _  1 


cot  X  = =  —  =  —  1. 

tan  X     —  1 


13.  Given  tana;=  V3;  find  the 
other  functions  when  x  is  an  angle 
in  Quadrant  III. 

tana3=  VS. 


cot  X  =  - —  =  ^y/S. 


tancc 


V3 

sin  X 

cos  a; 

tan  X  X  cos  x  =  sin  x. 

Vs  cos  a;  =  sin  x. 

3  cos%  —  sin^a;  =  0 
cos^x  +  sin^a;  =  1 


4  cos^a; 


cos'^a; 


cos  X 


=  1 


±i 


The  angle  being  in  Quadrant  III. 
the  cosine  is  negative. 


.".  cos  X 

=  -i- 

sin  X 

=  Vi-(- 

-W 

=  V|  =  ± 

JV3. 

Sine  is 

negative. 

.*.  sin  X 

=  -jV3. 

sec  X 

1 

1 

-2. 

esc  X 

1 

=  -# 

-*V3 


14.  Given  sec  a;  =  +  7,  and  tan  x 
negative ;  find  the  other  functions 
of  x. 

X  must  be  in  Quadrant  IV. 

.•.  sine,  cosine,  tangent,  and  co- 
tangent will  be  negative,  and  cosine 
positive. 


TEACHEES     EDITION. 


59 


cos  X  = 


sec  a; 


sma; 


^        49         >'49 


CSC  a;  = 


tana;  = 


cot  a; 


=  -  f  V3, 

1 

1 

sin  a;      _ 

|V3 

=  -tW3. 

sin  X      - 

-|V3 

cos  X 

1 

=  -  4  V3. 

1 
tan  X 

1 
4V3 

=  i^^V3. 

15.  Given  cot  a;  =  —  3;  find  all 
the  possible  values  of  the  other 
functions. 

By  [3]  tan  x  =  —  ^,  and  may  be 
in  Quadrant  II.  or  IV. 

By  [1], 

sin^a;  =  1  —  cos'^x. 


sin  X  = 

Vi 

—  cos^a;. 

By  [2], 

1 

Vi 

—  coa^a; 

3 

( 

308  a; 

1_ 

1- 

cos^a; 

y 

cos-'a; 

cos^a;  = 

9  —  9  cos^a;. 

cos^a;  = 

9 

lo' 

cos  X  = 

Vio 

and 

is  —  in 

Quadrant  II.,  +  in 

IV 

By  [1], 

sia  x  =  \l =  \  — 

\       in      \in 


10      ^10 
=  i^7VT0, 
and  is  +  in  Quadrant  II.,  —  in  IV. 

Vio 


sec  a; 


=  iVio. 


CSC  a;  =  VTO, 

16.  What  functions  of  an  angle 
of  a  triangle  may  be  negative  ?  In 
what  case  are  they  negative  ? 

When  an  angle  of  a  triangle  is 
acute,  its  functions  are  all  positive. 
When  an  angle  is  obtuse,  its  func- 
tions are  those  of  an  angle  in  Quad- 
rant II. 

.•.  sine  and  cosecant  are  positive, 
and  cosine,  tangent,  cotangent,  and 
secant  are  negative. 

17.  What  functions  of  an  angle 
of  a  triangle  determine  the  angle, 
and  what  functions  fail  to  do  so  ? 

The  sine  and  cosecant  being  posi- 
tive in  the  first  and  second  quad- 
rant, leave  it  doubtful  whether  the 
angle  is  obtuse  or  acute  ;  but  the 
other  functions,  if  positive,  deter- 
mine an  angle  in  the  first  quadrant, 
that  is  to  say,  an  acute  angle  ;  if 
negative,  an  angle  in  the  second 
quadrant,  an  obtuse  angle. 

18.  Why  may  cot  360°  be  con- 
sidered equal  either  to   -I-  oo  or  to 

-oo? 

The  nearer  an  acute  angle  is  to 
0°,  the  greater  the  positive  value  of 
its  cotangent ;  and  the  nearer  an 
angle  is  to  360°,  the   greater  the 


60 


TRIGONOMETRY. 


negative  value  of  its  cotangent. 
When  the  angle  is  0°  or  360°,  co- 
tangent is  parallel  to  the  horizontal 
diameter  and  cannot  meet  it.  But 
cotangent  360°  may  be  regarded 
as  extending  either  in  the  positive 
or  in  the  negative  direction ;  and 
hence  either  +  oo  or  —  oo  . 

19.  Obtain  by  means  of  Formu- 
las [l]-[3]  the  other  functions  of 
the  angles  given : 

(i.)  tan    90°  =  oo. 

(ii.)  cos  180°  =  - 1. 

(iii.)  cot  270°  =  0. 

(iv.)  esc  360°  =  -  00. 

(i-) 

tan90°  =  oo=l. 
0 


cot  90°  =  ~ 

CO 


sin  90° 


cos  90°      0 

cos  90° -0  sin  90°  =  0. 

cos2  90°  +  sin^  90°  =  1. 

sin2  90°  =  1. 

sin  90°  =  1. 

(ii.) 

cos  180°  =  -1. 

sin2 180°  +  cos2 180°  =  1. 

sin2 180°  +  1  =  1. 

sin  180°  =  0. 

sin  180°        0 


tan  180° 


cot  180°  = 


cos  180°      -  1 
=  -0. 

cos  180°  _  -  1 
sin  180°        0 


(ill-) 
cot  270°  =  0. 

tan  270°  =  -  =  oo. 
0 


cos  270° 


0. 


sm  270° 

cos  270°  =  0  sin  270°  =  0. 

sin2  270°-hcos2  270°  =  l. 

sin2  270°  +  0  =  l. 

sin2  270°  =  l. 

sin  270°  =  -  1. 

(IV.) 

CSC  360°  =  -  oo. 

sin  360°  =  —  =  -  0. 

—  00 

sin2  360°  +  cos2  360°  =  l. 

cos2  360°  =  l. 

cos  360°  =  1. 

tan  360°  =  ^  =  -  0. 


cot  360°  =  -!-  =  -  00. 
-0 


20.  Find  the  values  of  sin  450°, 
tan  540°,  cos  630°,  cot  720°,  sin  810°, 
CSC  900°. 

sin  450°  =  sin  (360°  +  90°) 

=  sin  90° 

=  1. 
tan  540°  =  tan  (360°  +  180°) 

=  tan  180° 

=  0. 
cos  630°  =  cos  (360°  +  270°) 

=  cos  270° 

=  0. 


TEACHERS     EDITION. 


61 


cot  720°  =  cot  (360°  +  360°) 
=  cot  360° 

=  00. 

sin  810°  =  sin  (2  x  360°  +  90°) 
=  sin  90° 
=  1. 

esc  900°  =  esc  (2  x  360° +180°) 
=  CSC  180° 

=  00. 

21.  For  what  angle  in  each 
quadrant  are  the  absolute  values  of 
the  sine  and  cosine  alike  ? 

The  sine  and  cosine  of  45°  are 
equal  in  absolute  value.  Corre- 
sponding to  the  angle  of  45°  in  the 
first  quadrant  are  the  angles  (90°  + 
45°),  (180° +  45°),  (270° +  45°)  in 
the  second,  third,  and  fourth  quad- 
rants. Hence  the  sines  and  cosines 
of  45°,  135°,  225°,  315°,  etc.,  are  all 
equal  in  absolute  value. 

22.  Compute  the  value  of 

a  sin  0°  +  5  cos  90°  -  c  tan  180°. 
sin  0°     =  0. 
cos  90°   =  0. 
tan  180°  =  0. 


Substituting, 

axO  +  6xO-cxO=0. 

23.  Compute  the  value  of 

a  cos  90°  -  h  tan  180°  +  c  cot  90°. 

cos  90°    =  0. 

tan  180°  =  0. 

cot  90°    =0. 
Substituting, 

ax0-6x0  +  cx0  =  0. 

24.  Compute  the  value  of 

a  sin  90°  -  5  cos  360° 

+  (a  -  6)  cos  180°. 

sin  90°    =  1. 
cos  360°  =  1. 
cos  180°  =  -  1.    ■ 
Substituting, 

axl-6xl+(a-5)x-l  =  0. 

25.  Compute  the  value  of 
(a2-52)cos  360°-  4  ah  sin  270°. 

cos  360°  =  1. 
sin  270°  =  -  1. 
Substituting, 

(a2-52)xl-4a&X-l 
=  a2-62  +  4a5. 


Exercise  X.     Page  41. 


2.  Express  sin  172°  in  terms  of 
the  functions  of  angles  less  than 
45°. 

sin  172°  =  sin  (180°  -  8°) 

=  sin  8°. 

3.  Express  cos  100°  in  terms  of 
the  functions  of  angles  less  than 
45°. 


cos  100°  =  cos  (90° +  10°) 
=  -  sin  10°. 


4.  Express  tan  125°  in  terms  of 
the  functions  of  angles  less  than 
45°. 

tan  125°  =  tan  (90°  -  35°) 

=  -  cot  35°. 


62 


TRIGONOMETRY. 


5.  Express  cot  91°  in  terms  of 
the  functions  of  angles  less  than 
45°. 

cot  91°  =  cot  (90°  +  1°) 
=  -  tan  1°. 

6.  Express  sec  110°  in  terms  of 
the  functions  of  angles  less  than 
45°. 

sec  110°  =  sec  (90°  +  20°) 

=  -  CSC  20°. 

7.  Express  esc  157°  in  terms  of 
the  functions  of  angles  less  than 
45°. 

CSC  157°  =  esc  (180° -23°) 

=  CSC  23°. 

8.  Express  sin  204°  in  terms  of 
the   functions    of  angles  less   than 

45°. 

sin  204°  =  sin  (180°  +  24°) 

=  -  sin  24°. 

9.  Express  cos  359°  in  terms  of 
the   functions    of  angles  less   than 

45°. 

cos  359°  =  cos  (360°  -  1°) 

=  cos  1°. 

10.  Express  tan  300°  in  terms  of 
the   functions   of  angles   less  than 

45°. 

tan  300°  =  tan  (270° +  30°) 

=  -  cot  30°. 

11.  Express  cot  264°  in  terms  of 
the  functions   of  angles   less   than 

45°. 

cot  264°  =  cot  (270° -6°) 

=  tan  6°. 

12.  Express  sec  244°  in  terms  of 
the   functions   of  angles  less  than 

45°. 

sec  244°  =  sec  (270° -26°) 

=  -csc26°. 


13.  Express  esc  271°  in  terms  of 
the  functions  of  angles  less  than 
45°. 

CSC  271°  =  CSC  (270°  -  1°) 
=  -secl°. 

14.  Express  sin  163°  49''  in  terms 
of  the  functions  of  angles  less  than 
45°. 

sin  163°  49^=  sin  (180°- 16°  110 
=  sin  16°  IV. 

15.  Express  cos  195°  33^  in  terms 
of  the  functions  of  angles  less  than 
45°. 

cos  195°  33^=  cos  (180°+ 15°  33^ 
=  -  cos  15°  33^ 

16.  Express  tan  269°  15^  in  terms 
of  the  functions  of  angles  less  than 
45°. 

tan  269°  15^=  tan  (270°- 450 
=  cot45^ 

17.  Express  cot  139°  17^  in  terms 
of  the  functions  of  angles  less  than 
45°. 

cot  139°  17^=  cot  (180°-  40°  430 
=  -  cot  40°  43^ 

18.  Express  sec  299°  45''  in  terms 
of  the  functions  of  angles  less  than 
45°. 

sec  299°  45^=  sec  (270°+  29°  450 
=  CSC  29°  45^. 

19.  Express  esc  92°  25''  in  terms 
of  the  functions  of  angles  less  than 
45°. 

esc  92°  25^=  CSC  (90°+  2°  250 
=  sec  2°  25^ 

20.  Express  all  the  functions  of 
—  75°  in  terms  of  those  of  positive 
angles  less  than  45°. 


TEACHERS     EDITION, 


63 


sin  (-  75°)  =  sin  (270°  +  15°) 

=  -  cos  15°. 
cos  (-  75°)  =  cos  (270°  +  15°) 

=  sin  15°. 
tan(-  75°)  =  tan  (270°  +  15°) 

=  -  cot  15°. 
cot  (-  75°)  =  cot  (270°  +  15°) 

=  -  tan  15°. 


21.   Express  all  the  functions  of 
—  127°  in  terms  of  those  of  positive 
angles  less  than  45°. 
sin  (-  127°)  =  sin  (270°  -  37°) 

=  -  cos  37°. 
cos  (-  127°)  =  cos  (270°  -  37°) 

=  -  sin  37°. 
tan  (-  127°)  =  tan  (270°  -  37°) 

=  cot  37°. 
cot  (- 127°)  =  cot  (270°  -  37°) 

=  tan  37°. 


22.   Express  all  the  functions  of 
—  200°  in  terms  of  those  of  positive 
angles  less  than  45°. 
sin  (-  200°)  =  sin  (180°  -  20°) 

=  sin  20°. 
cos  (-  200°)  =  cos  (180°  -  20°) 

=  -  cos  20°. 
tan  (-  200°)  =  tan  (180°  -  20°) 

=  -tan20°. 
cot  (-  200°)  =  cot  (180°  -  20°) 

=  -  cot  20°. 


23.  Express  all  the  functions  of 
—  345°  in  terms  of  those  of  positive 
angles  less  than  45°. 

sin  (-  345°)  =  sin  15°,  etc. 


24.  Express  all  the  functions  of 
—  52°  37''  in  terms  of  those  of  posi- 
tive angles  less  than  45°. 

sin  (-  52°  370  =  sin  (270°-  37°  23^ 

=  -  cos  37°  23^ 
cos  (-  52°  370  =  cos  (270°+  37°  23^ 

=  sin  37°  23^ 
tan  (-  52°  370  =  tan  (270°+  37°  230 

=  -  cot  37°  23^. 
cot  (-  52°  370  =  cot  (270°+  37°  230 

=  -tan37°23^ 

25.  Express  all  the  functions  of 
— 196°  54-^  in  terms  of  those  of  posi- 
tive angles  less  than  45°. 

sin  (- 196°  540  =  sin  (180°-16°540 
=  sin  16°  54^ 

cos  (- 196°  540  =  cos  (180°-16°  540 
=  -  cos  16°  54^, 

tan  (- 196°  540  =  tan  (180°-16°  540 
=  -tanl6°54^ 

cot  (- 196°  540  =  cot  (180°-16°  540 
=  -cotl6°54^ 

26.  Find  the  functions  of  120°. 
sin  120°  =  sin  (90°  +  30°)  =  cos  30° 

=  |V3. 
cos  120°  =  cos  (90°  +  30°) 
=  -  sin  30°  =  -  J. 

tan  120°  =  tan  (90° +  30°) 

=  - tan  30°  = -Vs. 

cot  120°  =  cot  (90°  +  30°) 

=  -cot30°  =  -^V3. 

sec  120°  =  -  2. 

CSC  120°  =  f  Vs. 


64 


TRIGONOMETRY. 


27.    Find  the  functions  of  135°. 
sin  135°  =  sin  (90°  +  45°) 


cos  45°  =  I V2. 


cos  135°  =  cos  (90°  +  45°) 

=  -sin45°  =  - jVi 
sin  135° 


tan  135°  = 


cos  135° 

cotl35°  =  ^"^^^^°  =  -l. 
sin  135° 


1 


sec  135°  = ^—  =  -  \/2. 

CSC  135°  = 


cos  135° 

1  V2. 


sin  135° 

28.   Find  the  functions  of  150°. 
sin  150°  =  sin  (180°  -  30°) 

=  sin  30°  =  J. 
cos  150°  =  COS  (180°  -  30°) 

=  -cos30°  =  -jV3. 
tanl50°  =  tan(180°-30°) 

sin  30° 


=  -tan30°  = 


-  cos  30° 
=  -iV3. 

cot  150°  =  cot  (180°  -  30°) 

-  cos  30° 


=  -  cot  30°  = 


sin  30° 


=  -V3. 
sec  150°  =  sec  (180°  -  30°) 
=  -  sec  30°  = 


-  cos  30° 


=  -fV3. 
CSC  150°  =  CSC  (180°  -  30°) 

=  CSC  30°  = 


sin  30° 


=  2. 


29.  Find  the  functions  of  210°. 
sin  210°  =  sin  (180°  +  30°) 

=  -  sin  30°  =  -  |. 
cos  210°  =  cos  (180°  +  30°) 

=  -cos30°  =  -A\/3. 
tan  210°  =  tan  (180° +  30°) 

=  tan30°  =  J\/3. 
cot  210°  =  cot  (180°  +  30°) 

=  cot  30°  =  V3. 

30.  Find  the  functions  of  225°. 
sin  225°  =  sin  (180°  +  45°) 

=  -sin45°  =  -|\/2. 
cos  225°  =  cos  (180°  +  45°) 

=  -cos45°  =  - J\/2. 
tan  225°  =  tan  (180° +  45°) 

=  tan  45°  =  1. 
cot  225°  =  cot  (180°  +  45°) 

=  cot45°  =  l. 

31.  Find  the  functions  of  240°. 
sin  240°=  sin  (270°  -  30°) 

=  -cos30°  =  -jV3. 
cos  240°  =  cos  (270°  -  30°) 

=  -  sin  30°  =  -h 
tan  240°  =  tan  (270° -30°) 

=  cot  30°  =  Vs. 
cot  240°  =  cot  (270°  -  30°) 

=  tan  30°  =  ^  Vs. 

32.  Find  the  functions  of  300°. 
sin  300°  =  sin  (270°  +  30°) 

=  -cos30°  =  -jV3. 

cos  300°  =  cos  (270°  +  30°) 
1 

2- 


TEACHERS     EDITION. 


65 


tan  300°=  tan  (270° +  30°) 

=  -  cot  30°  =  -  V3. 

cot  300°  =  cot  (270°  +  30°) 

=  -tan30°=-jV3. 

33.   Find  the  functions  of  -  30°. 


sin  -  30°  =  -  sin  30°  =  -  i. 

cos  -  30°  = 

tan  -  30°  = 

cot  -  30°  = 

sec  -  30°  = 

esc  -  30°  = 


cos  30°  =  J  Vs. 
-tan30°  =  -^V'3. 

-  cot  30°  =  -  Va. 
sec  30°  =  |V3. 

-  CSC  30°  =  -  2. 


34,   Find  the  functions  of  -225°. 

-  225°  =  90°  +  45°. 

sin  -  225°  =  sin  (90°  +  45°) 

=  cos  45°  =  \V2. 
cos  -  225°  =  cos  (90°  +  45°) 

=  -  sin  45°  =  ~  I V2. 
tan  -  225°  =  tan  (90°  +  45°) 

=  -  cot  45°  =  -  1. 
cot  -  225°  =  cot  (90°  +  45°) 

=  -  tan  45°  =  -  1. 
1 


sec  -  225°  = 


cos  (90°  +  45°) 
=  -V2. 

1  =V2. 


CSC  -  225°  =  — 

sin  (90° +  45°) 

35.  Given  sin  a;  = —Vl,  and  cos  a? 
negative ;  find  the  other  functions 
of  X,  and  the  value  of  x. 

Since  sin  45°  ==  V|,  and  the  signs 
of  both  the  sine  and  cosine  are  neg- 
ative, the  angle  must  be  in  Quadrant 
III.,  and  must  be,  therefore, 

180°  +  45°  =  225°. 

Then  cos45°=\/J. 

Hence  cos  (180°  +  45°)  =  -  VJ. 


sin  225^ 


-T  ^^    )  — 

cos  225° 

= 

-x4 

(180° 

+  45°)  = 

1 

tan  225° 

225° 

1 

1 

COS  225 

°  -vi 

=  - V2. 

CSC 

225° 

1 

1 

sm  225°     _VJ 
=  -V2. 

36.  Given  cot  x  =  —  \/3,  and  x  in 
Quadrant  II.  ;  find  the  other  func- 
tions of  X,  and  the  value  of  x. 

Since  cot30°=\/3,  and  the  sign 
is  negative,  the  angle  is  in  Quad- 
rant II. 

tan  X  =  ■ =  ■ =  —  jVS. 

cot  a;     —  V3 

!ilLiE  =  -|V3. 

cos  X 

sin  x  =  —  I  VS  cos  X. 
sin^a;  =  ^  cos^a:. 
But  sin'^aj  +  cos^a?  =  1, 
.'.   J  cos^a;  +  cos^aj  =  1 ; 


and 


and 


cos^a; 


,*.  cos  x=  ^v3 

; 

•     2            1 

4 

.'.  sin  x  =  — 

2 

sec  X  = 

=  fV3. 

cos  a? 

esc  X  =  - — 

=  2. 

sm  X 


66 


TKIGONOMETBY. 


37.    Find  the  functions  of  3540°. 
3540«  =  9  X  360°  +  300°. 

sin  300°  =  sin  {S&)P  -  m^} 

=  sin  60°  =  - 1  VS. 
cos  aOO^'  =  cos  (360°  -  60°) 


tan  300^  = 


cot  300° 


cos  60^  =  i. 
2 

smg00^^-|V3 

cos  SOO'*  i 

1  1 


secSCKT 
esemF  = 


tan  300°      _v^ 

-|V3. 

1 


€Os3(X>'^     i 


=  -  =  2L 


sin  300*^ 


^V3 


Sa  What  angles  less  ^an  360°' 
have  a  sine  equal  to  —  J  ?  a  tangent 
equal  to  —  V3  ? 

(i.)  ^aee  sin  30^=  J  and  the  sign 
is  negative,  tlie  angle  nrast  be  in 
Qnadraat  III.  or  IV.,  and  mnst  be 
therefore  180°  +  30°  =  210°,  or  360° 
-30°  =  330°. 

(ii.)  Since  tan  60°  =V3  and  the 
sign  is  negative,  the  angle  must  be 
in  Quadrant  II.  or  IV.,  and  must 
be  therefore  180°-60°  =  12CP,  or 
360° -60°  =  300°. 

39.  Which  of  the  angles  men- 
tioned in  Examples  27-34  have  a 
cosine  equal  to  —  VJ?  a  cotangent 
equal  to  —  V3  ? 

(i.)  Since  cos45°=V^  and  the 
sign  is  negative,  the  angle  must  be 


in  Quadrant  II.  or  III.,  and  must 
be  therefore  180°  -  45°  =  135°,  or 
180°  +  45°  =  225°.  Also,  the  func- 
tions of  —  225°  are  the  same  as  the 
functions  of  360°  -  225°  =  135°. 
Hence  the  angles  are  135°,  225°,  or 
-225°. 

(ii.)  Since  cot  30°  =  V3  and  the 
sign  is  negative,  the  angle  must  be 
in  Quadrant  II.  or  IV.,  and  must 
be  therefore  180°  -  30°  =  150°,  or 
360°  -  30°  =  330°,  or  -  30°.  Hence 
the  angles  are  150°  or  —  30°. 

40.  What  values  of  »■  between  O*' 
and  720°  will  satisfy  the  equation 
sin  X  =  4r  2? 

Since  sin  30°  =  -|  and  the  sign  is 
positive,,  the  angle  must  be  in  Quad- 
rant I.  or  II.,  and  must  be  therefore 
30°,  or  180°-  30°  =  150°,  the  first 
revolution.  In  the  second  revolu- 
tion, these  angles  must  be  increased 
by  360°.  Hence  the  angles  are  30°, 
150°,  390°.  and  510°. 


41.  In  each  of  the  following  eases 
find  the  other  angle  between  0°  and 
360°  for  which  the  corresponding 
function  (sign  included)  has  the 
same  value :  sin  12°,  cos  26°,  tan  45°, 
cot  72°  ;  sin  191°,  cos  120°,  tan  244°, 
cot  357°. 

In  order  that  the  sign  shall  be 
the  same, 

sin  12°  must  be  in  Quadrant  II. 

=  sin(lSO^-    12°)  =  sin  168°. 
cos  26°  miist  be  in  Quadrant  IV. 

=  cos  (360°  -  26°)  =  cos  334°. 
tan  45°  must  be  in  Quadrant  III. 

=  tan  (180°  +  45°)  =  taa  225^ 


TEACHERS     EDITION. 


67 


cot  72°  must  be  in  Quadrant  III. 

=  cot  (180°  +  72°)  =  cot  252°. 
sin  191°  must  be  in  Quadrant  IV. 

=  sin  (360°  -  11°)  =  sin  349°. 
cos  120°  must  be  in  Quadrant  III. 

=  cos  (180°  +  60°)  =  cos  240°. 
tan  244°  must  be  in  Quadrant  I. 

=  tan  (244°  -  180°)  =  tan  64°. 
cot  357°  must  be  in  Quadrant  II. 

=  cot  (357°  -  180°)  =  cot  177°. 

42.   Given   tan  238°  =  1.6;    find 
sin  122°. 

tan  238°  =  tan  (180°  +  58°) 

-tan  58°. 
sin  122°  =  sin  (180°  -  58°) 
,  =  sin  58°. 
But  tan  238°  =  1.6. 
.-.      tan   58°  =  1.6. 

sin  58° 


tan  58°    = 
1 


cos  58° 
sin  58° 


Vi-sin258° 
2.56  -  2.56  sin2  58°  =  sin2  58°. 
3.56  sin2  58°  =  2.56. 

sm  58°  =  -\/ ■ 

^'3.56 

=  0.848. 

43.   Given  cos  333°  =  0.89  ;    find 
tan  117°. 

cos  333°  =  0.89. 

=  cos  (270°  +  63°) 

=  sin  63° 

=  tan  (180° -63°) 

=  -  tan  63°. 
Bin2  63°  +  cos2  63°  =  1. 
(0.89)2  +  cos2  63°  =  1. 


cos2  63°  =  0.2079. 

cos  63°  =  0.456. 

tan  63°=     '^^  ^^° 
cos  63° 

0.89 

- 1.952 

0.456 

44.  Simplify  the  expression 

a  cos  (90°  -x)+h  cos  (90°  +  x) 
=  a  sin  X  +  b{—  sin  x) 
=  sin  x{a  —  h). 

45.  Simplify  the  expression 
m  (cos  90°  -  a;)  sin  (90°  -x). 
cos  (90°  —  x)  =  sin  x. 

sin  (90°  —  x)  =  cos  x. 

.'.  the  expression  =msmx  cos  x. 

46.  Simplify  the  expression 
(a -J)  tan  (90° -k) 

+  (a  +  &)cot(90°+4 
tan  (90°  -  x)  =  cot  x. 
cot  (90°  +  x)  =  —  tan  x. 
.'.  the  expression  equals 
(a  —  b)  cot  X  —  {a  +  b)  tan  x. 

47.  Simplify  the  expression 
a2  +  62_2a6cos(180°-a;) 

=  a^  -\-  b^  —  2ab  {—  cos  x) 
=  a^  +  b^  +  2ab  cos  x. 

48.  Simplify  the  expression 
sin  (90°  +  x)  sin  (180°  +  x) 

+  cos  (90°  +  x)  cos  (180°  -  x) 
=  (cos  x)  (—  sin  a?)+(— sin  x)(—  cos  x) 
=  —  sin  cos  X  +  sin  cos  x 
=  0. 

49.  Simplify  the  expression 
cos  (180°  +  x)  cos  (270°  -  y) 

-  sin  (180°  +  x)  sin  (270°  -  y). 


68 


TRIGONOMETRY. 


COS  (180°  +  x)  =  —  cos  X. 
cos  (270°  —y)  =  —  sin  y. 
sin  (180°  +  x)^  —  sin  x. 
sin  (270°  —  y)  =  —  cos  y. 
Hence  the  expression 
=  cos  X  sin  y  —  sin  x  cos  3/. 

50.  Simplify  the  expression 

tan  x  +  tan  (—  y)  —  tan  (180°  —  y). 
tan  (—  2/)  =  —  tan  1/. 
-tan  (180° -3/)  =  tan  2/. 
Hence  the  expression  =  tan  x. 

51.  For  what  values  of  x  is  the 
expression  sin  x  +  cos  x  positive, 
and  for  what  values  negative  ?  Rep- 
resent the  result  by  a  drawing  in 
which  the  sectors  corresponding  to 
the  negative  values  are  shaded. 

If  X  be  any  angle  in  Quadrant  I., 
sin  X  +  cos  X  must  be  positive  since 
both  the  sine  and  cosine  are  posi- 
tive. In  Quadrant  II.  the  sine  is 
positive  and  cosine  negative ;  hence, 
so  long  as  the  sine  is  greater  than, 
or  equal  to,  the  cosine,  the  expres- 
sion sin  a? -f  cos  re  is  positive;  but 
after  passing  the  middle  of  Quad- 
rant II.,  viz.,  135°,  the  cosine  of 
Zx  is  greater  than  sine,  and  the  ex- 
pression is  negative.  In  Quadrant 
III.  both  sine  and  cosine  are  nega- 
tive, and  hence  their  sum  must  be 
negative.  In  Quadrant  IV.  the 
sine  is  negative  and  cosine  posi- 
tive. The  sine  and  cosine  are  equal 
at  315°,  after  which  the  cosine  is 
greater  than  sine.  Hence  the  ex- 
pression sin  X  +  cos  X  is  negative 
from  135°  to  315°,  and  positive  be- 
tween 0°  and  135°,  and  315°  and 
360°. 


52.  Answer  the  question  of  last 
example  for  sin  x  —  cos  x. 

As  X  increases  from  0°  to  45°,  the 
sine  increases  in  value,  and  cosine 
decreases,  until  at  45°  sine  =  cosine. 
Hence  up  to  this  point  sin  .r  — cos  it 
is  negative.  For  the  remainder  of 
Quadrant  I.  sine  is  greater  than  co- 
sine, and  consequently  the  expres- 
sion sin  X  —  cos  X  is  positive.  In 
Quadrant  II.  sine  is  positive  and 
cosine  negative,  so  the  expression 
sin  X  —  cos  X  is  uniformly  positive. 
In  Quadrant  III.  sine  is  negative 
and  cosine  negative  ;  hence,  so  long 
as  sine  is  less  than  cosine,  the  ex- 
pression is  positive,  viz.,  to  225°; 
after  that  point,  sine  is  greater  than 
cosine,  and  sin  x  —  cos  x  is  negative. 
In  Quadrant  IV.  sine  is  negative 
and  cosine  positive  :  therefore  sin  x 
—  cos  a;  is  uniformly  negative.  The 
expression  is,  then,  negative  be- 
tween 0°  and  45°,  and  225°  and 
360°;  positive  between  45°  and 
225°. 

53.  Find  the  functions  of 

(x  —  90°)  in  terms  of  the  functions 

^^^-     X-  90°  =  360°  -  (90°  -  x) 

=  270°  -f  X. 
sin  {x  -  90°)  -  sin  (270°  +  x) 

=  —  cos  X. 
cos  {x  -  90°)  -  cos  (270°  -1-  x) 

=  sin  X. 
tan  {x  -  90°)  -  tan  (270°  -f  x) 

=  —  cot  X. 
cot  {x  -  90°)  =  cot  (270°  +  x) 

=  —  tan  X. 

54.  Find  the  functions  of 

{x  —  180°)  in  terms  of  the  functions 
of  cc. 


TEACHERS     EDITION. 


69 


X  -  180°  =  360°  -  (180°  -  X) 
=  180°  +  X. 

sin  {x  -  180°)  =  sin  (180°  +  x) 

=  —  sin  X. 
cos  {x  -  180°)  =  cos  (180°  +  x) 


=  —  cos  X. 
tan  (a;  -  180°)  =  tan  (180°  -  x) 

=  tan  X. 
cot  {x  -  180°)  =  cot  (180°  +  x) 

=  cot  X. 


Exercise  XI.     Page  48. 


1.   Find  the  value  of  sin  {x  +  y) 
and  cos  {x  +  y)  when  sin  a?  =  |,  cos  x 


=  f,  sin 2/ -tV.  cosy 


12 
13- 


sin  (x  +  y)  =  sin  x  cos  y  +  cos  x  sin  y 

+ 


'5^13 


36     20  _  56^ 
65     65     65* 


5^13 


cos  (a?  +  3/)  =  cos  X  cos :«  —  sin  x  sin  1/ 


"'5^13)      (5^13 


48 
65 


15 
65 


33 
65" 


2.  Find  sin  (90°  —  y)  and  cos 
(90° -y)  by  making  a;  =  90°  in 
Formulas  [8]  and  [9]. 

sin  (90°  -  y) 

=  sin  90°  cos  y  —  cos  90°  sin  y. 
sin  90°  =  1.     cos  90°  =  0. 
.-.  sin  (90°  -  y) 

=  (1 X  cosy)  —  (0  X  sin  y) 

=  cos  y. 
cos  (90°  -  y) 

=  cos  90°  cos  y  +  sin  90°  sin  y 

=  (0  X  cos  y)  +  (1  X  sin  y) 

=  sin  y. 

3.  Find,  by  Formulas  [4H11], 
the  first  four  functions  of  90°  +  y. 


sin  (90°  +  y) 

=  sin  90°  cos  y  +  cos  90°  siny 
=  (1  Xcosy)  +  (Oxsiny) 
=  cos  y. 

cos  (90°  +  y) 

=  cos  90°  cos  y  —  sin  90°  sin  y 
=  (0  X  cos  y)  —  (1  X  sin  y) 
=  —  sin  y. 

(tan  90°  +  y) 

cos  y  , 

= -, — ^  =  —  cot  y. 

sin  y 

cot  (90°  +  y) 

sin  V  , 

= ^  =  —  tan  y. 

cosy 

4.   Find,   by  Formulas   [4]-[ll], 
the  first  four  functions  of  180°  —  y. 

sin  (180°  -  y) 

=  sin  180°  cosy  — cos  180°  siny 
=  (0  X  cos  y)  —  (—  1  X  sin  y) 
=  sin  y, 

cos  (180°  -  y) 

=  cos  180°  cos  y  +  sin  180°  sin  y 
=  (—  1  X  cos  y)  +  (0  X  sin  y) 
=  —  cosy. 

tan  (180°  -  y) 

= ^  =  —  tan  y. 

cosy 

cot  (180°  -  y) 

cos  V  , 

= ; — ^  =  —  cot  y. 

sin  y 


70 


TRIGONOMETEY. 


5.  Find,  by  Formulas   [4]-[ll], 
the  first  four  functions  of  180°  +  y. 

sin  (180°  +  y) 

=  sin  180°  cos  y  +  cos  180°  sin  y 
=  (0  X  cos  y)  +  (-  1  X  sin  y) 
=  —  sin  y. 

cos  (180°  +  y) 

=  cos  180°  cos  y  —  sin  180°  sin  y 
=  (—  1  X  cos  2/)  —  (0  X  sin  y) 
=  —  cos  y. 

tan  (180°  +  y) 

—  sin  v      , 

= ^  =  tan  y. 

—  cosy 

cot  (180°  +  y) 

—  cos  y  , 

= : — ^  =  cot  y. 

—  sin  y 

6.  Find,   by  Formulas  [4]-[ll], 
the  first  four  functions  of  270°  — y. 

sin  (270°  -  y) 

=  sin  270°  cos  y  —  cos  270°  sin  y 
=  (—  1  X  cos  2/)  —  (0  X  sin  y) 
=  —  cos  y. 

cos  (270°  -  y) 

=  cos  270°  cos  y  +  sin  270°  sin  y 
=  (0  X  cos  2/)  +  (-  1 X  sin  y) 
=  —  sin  y. 

tan  (270° -2/) 

—  cos  v         . 

=  — ; — ^  =  cot  y. 

—  sin  y 

cot  (270°  -  y) 

=  :i^iM  =  tan2/. 

—  cos  2/ 

7.  Find,  by  Formulas   [4]-[ll], 
the  first  four  functions  of  270°  +  y. 

sin  (270°  +  y) 

=  sin  270°  cos  y  +  cos  270°  sin  y 
=  (-  1  X  cos  2/)  +  (0  X  sin  y) 
=  —  cos  y. 


cos  (270°  +  y) 

=  cos  270°  cos  y  —  sin  270°  sin  y 
=  (0  X  cos  2/)  —  (—  1  X  sin  y) 
=  sin  y. 

tan  (270°  +  y) 

—  cosy  , 

=  — : ^  =  —  cot  y. 

sm  y 

cot  (270°  +  y) 

sin  V  i. 

= "^  =  —  tan  y. 

—  cos  y 

8.  Find,  by  Formulas    [4]-[ll], 
the  first  four  functions  of  360°  —  y. 

sin  (360°  -  y) 

=  sin  360°  cos  y  -  cos  360°  sin  y 
=  (0  X  cos  2/)  —  (1  X  sin  y) 
=  —  sin  y. 

cos  (360°  -  y) 

=  cos  360°  cos  y  +  sin  360°  sin  y 
=  (1  X  cos  2/)  +  (0  X  sin  y) 
=  cos  2/. 

tan  (360°  -  y) 

—  sin  v  , 

= -^  =  —  tan  y. 

cosy 

cot  (360° -2/) 

cos  y  , 

=  — r-"^  =  —  cot  y. 

—  sin  2/ 

9.  Find,   by  Formulas   [4]-[ll], 
the  first  four  functions  of  360°  +  y. 

sin  (360°  +  y) 

=  sin  360°  cos  y  +  cos  360°  sin  y 
=  (0  X  cos  2/)  +  (1  X  sin  y) 
=  sin  2/. 

cos  (360°  +  y) 

=  cos  360°  cos  y  —  sin  360°  sin  y 

=  (1  X  cos  y)  —  {Ox  sin  y) 

=  cos  y. 

tan  (360° +2/) 
_  sm_y  ^  ^^^  ^^ 

cos  2/ 


TEACHERS     EDITION. 


1 


cot  (360°  +  y) 

cos  y  , 

=  — — ^  =  cot  y. 
siny 

10.    Find,  by  Formulas  [4]-[ll], 
the  first  four  functions  of  x  —  90°. 

sin  {x  -  90°) 

=  sin  X  cos  90°  —  cos  x  sin  90° 
=  (0  X  sin  a;)  —  (1  X  cos  x) 
=  —  cos  X. 

cos  (x  -  90°) 

=  cos  X  cos  90°  +  sin  x  sin  90° 
=  (0  X  cos  x)  +{lx  sin  x) 
=  sin  X. 

tan  (a;  -  90°) 
—  cos  a; 


=  —  cot  X. 


sin  X 
cot  {x  -  90°) 
_    sin  X 
—  cosx 


=  —  tan  X. 


11.    Find,  by  Formulas  [4]-[ll], 
the  first  four  functions  of  a;  —  180°. 

sin  {x  - 180°) 

=  sin  x  cos  180°  —  cos  x  sin  180° 
=  sin  x  (—  1)  —  cos  a;  X  0 
=  —  sin  X. 

cos  (a;  - 180°) 

=  cos  X  cos  180°  +  sin  x  sin  180° 
=  cos  X  (—  1)  +  sin  a;  X  0 
=  —  cos  X. 

tan  (a;  - 180°) 
—  sin  X 


=  tan  X. 


—  cos  X 

cot  (a;  -  180°) 

—  cos  X 


cot  a;. 


sin  X 


12.   Find,  by  Formulas  [4]-[ll], 
the  first  four  functions  of  a;— 270°. 


sin  (a;  -  270°) 

=  sin  X  cos  270°  —  cos  x  sin  270° 
=  sin  a;  X  0  —  cos  a;  X  (—  1) 
=  cos  X. 

cos  (a;  -  270°) 

=  cos  X  cos  270°  +  sin  x  sin  270° 
=  cos  a;  X  0  +  sin  x  (—  1) 
=  —  sin  X. 

tan  (a; -270°) 
cos  X 


=  —  cot  X. 


—  sm  X 
cot  (a;  -  270°) 
_  —  sin  X  _ 
cos  X 


—  tan  a;. 


13.   Find,  by  Formulas  [4]-[ll], 
the  first  four  functions  of  —  y. 

sin  (0°  -  y) 

=  sin  0°  cos  y  —  cos  0°  sin  y 
=  (0  X  cos  y)  —  (1  X  sin  y) 
=  —  sin  y. 

cos  (0°  -  y) 

=  cos  0°  cos  y  +  sin  0°  sin  y 
=  (1  X  cos  2/)  +  (0  X  sin  y) 
=  cos  y. 

tan(0°-3/) 

—  sin  V  . 

= ^  =  —  tan  y. 

cosy 

cot(0°-y) 

cos  y  , 

=  — -"^  =  —  cot  v. 


—  siny 


14.   Find,  by  Formulas  [4]-[ll], 
the  first  four  functions  of  45°  —  y. 

sin  (45°  —  y) 

=  sin  45°  cos  y  —  cos  45°  sin  y 
=  2^V2  cos  y  —  ^\/2  siny 
=  ^  V2  (cos  y  —  sin  y). 


72 


TRIGONOMETEY. 


COS  (45°  —  y) 

=  cos  45°  cos  y  +  sin  45°  siny 

=  |V2  cos  2/  +  ^ y/2  sin  y 

=  \yf2  (cos  2/  +  sin  y). 
tan  (45°  -  y) 

_  cos  y  —  sin  y  _  1  —  tan  y 
cos  y  +  sin  3/      1  +  tan  y 
cot  (45°  -  y) 

_  cos  y  +  sin  y  _  cot  y  +  1 
cos  y  —  sin  y     cot  y  —  1 

15.   Find,  by  Formulas  [4]-[ll], 
the  first  four  functions  of  45°  +  y. 

sin  (45°  +  y) 

=  sin  45°  cos  y  +  cos  45°  sin  y 


=  iV2 


cosy  +  jVlsiny 


=  J  V2  (cos  y  +  sin  y). 
cos  (45°  +  y) 

=  cos  45°  cos  y  —  sin  45°  sin  y 
5=  J V2  cos  y  —  ^\/2  sin  y 
=  J  V2  (cos  y  —  sin  y). 

tan  (45°  +  y) 

_  cos  y  +  sin  y  _  1  +  tan  y 
cos  y  —  sin  y      1  —  tan  y 
cot  (45°  +  y) 

_  cos  y  —  sin  y  _  cot  y  —  1 
cos y  4- sin y     coty  +  1 

16.   Find,  by  Formulas  [4] -[11], 
the  first  four  functions  of  30°  +  y. 

sin  (30°  +  y) 

=  sin  30°  cos  y  +  cos  30°  sin  y 

=  \  (cos  y  +  \/3  sin  y). 
cos  (30°  +  y) 

=  cos  30°  cos  y  —  sin  30°  sin  y 

=  J  (  V3  cos  y  —  sin  y). 


tan  (30°  +  y) 

_  cos  y  +  \/3  sin  y  ^ 
Vo  cos  y  —  sin  y 
divide  each  term  by  VS  cos  y, 
_  ^  V3  +  tan  y 
1  —  jV3  tany 
cot  (30°  +  y) 

_  V3  cos  y  —  sin  y  . 
cosy  +  VSsiny 
divide  each  term  by  sin  y, 

=  "V^  cot  y  —  1 
coty  +  VS 

17.  Find,  by  Formulas  [4]-[ll]. 
the  first  four  functions  of  60°  —  y. 

sin  (60°  -  y) 

=  sin  60°  cos  y  —  cos  60°  sin  y 

—  K"^  cos  y  —  sin  y). 
cos  (60°  -  y) 

=  cos  60°  cos  y  +  sin  60°  sin  y 

=  2  (cos  y  +  \/3  sin  y). 
tan  (60°  -  y) 

_  Vs  cos  y  —  sin  y 
cos  y  +  V3  sin  y 

_  V3  —  tan  y 
1  +  VS  tan  y 
cot  (60° -y) 

_  cos  y  +  VB  sin  y 

VS  cos  y  —  sin  y 
_^V3coty  +  1 

cot  y  —  \  Vs 

18.  Find  sin  3  a;  in  terms  of  sin  x. 
sin  3x  =  sin  (2a;  +  x) 

=  sin 2 a;  cos X  + cos 2 a;  since. 


TEACHERS     EDITION. 


73 


sin  2x  =  2  sin  x  cos  x. 

cos  2  a;  =  cos^o;  —  sin^rc. 

Substituting, 

sin  3x  =  2  sin  re  cos^a; 

+  sin  X  cos^a;  —  sin^x 

=  3  sin  X  cos^aj  —  sin^a;. 
But  cos^a;  =  1  —  sin^a;. 
Substituting, 
sin  3  a;  =  3  sin  x  —  3  sin^x  —  sin'a; 

=  3  sin  x  —  4:  sin^a;. 

19.  Find  cos  3  a;  in  terms  of  cos  a;, 
cos  3x  =  cos  {2x  +  x) 

=  cos  2  X  cos  X  —  sin  2  x  sin  x. 
sin  2  a;  =  2  sin  x  cos  x. 
cos  2  a;  =  cos^a;  —  sin^a;. 
Substituting, 
cos  3  a;  ==  cos^a;  —  sin^a;  cos  x 

—  2  sin^x  cos  x 

=  cos^a;  —  3  sin'^a;  cos  x. 
But  sin%  =  1  —  cos^a;. 
Substituting, 
cos  3  a;  =  cos^a;  —  3  cos  a;  +  3  cos'a; 

=  4  cos%  —  3  cos  x. 

20.  Given  tan  |  a;  =  1 ;  find  cos  x. 

1 


tan 


cos  a; 
+  cosa; 


=V} 


cos  X 


1  = 


+  cosa; 
1  —cos a;  • 


1  +  cos  X 
1  +  cos  a;  =  1  —  cos  a;. 
2  cos  X  =  0. 
cos  x  =  0. 

21.  Given    cot  |  a;  =  V3  ;     find 
sin  a;. 


cot  Ja;='Y- 


+cosa; 


cos  a; 


V3 


i  +  cos  X. 


^l- 


3  = 


cos  a; 

1  +  cos  X 


1  —  cos  X 
3  —  3  cos  x  =  1  +  cos  X. 
—  4  cos  X  =  —  2. 


cosx  =  -• 
2 

sin^x  =  1  —  cos^x 


=  1- 


1^3 
4     4' 


sm  X 


-4-iVs. 


22.   Given  sin  x  =  0.2 ;  find  sin  ^  x 
and  cos^x. 

sin  X  =  0.2. 
cos^x  =  1  —  sin'^x 
=  1  -  0.04. 

cos  X  =  VOM. 


sin  ix  =  ^- 


—  cosx 


1-Va96 


1-oaVq 


=  0.10051 
cos  J  X  =  ^  - 


+  cos  X 


=^ 


I'l  +  0.4  VS 


=  0.99494. 


74 


TRIGONOMETRY. 


23.   Given  cos  a;  =  0.5  ;  find  cos  2  x 

and  tan  2  x. 

cos  2  .<;  =  cos'^a;  —  sin^a;. 


sin  a;     = 


W'-(iy-' 


Vs. 


cos  2a;  =  0.25 -0.75 

=  _  0.50  =  - 1. 
2 


tan  a;    = 
tan  2  a;  = 


sm  X 


COS  a;         ^ 
2  tan  a;        2\/3 


1  —  tan^a;      1  —  3 
=  -V3. 

24.   Given  tan  45°  =  1 ;    find  the 
functions  of  22°  30^ 


tana;    = 


sin  X 


cos  a; 
.'.  sin  X    =  cos  X. 
%va?-x  +  cos'^a;  =  1. 
sin^a;  +  sin'^a;  =  1. 
2sin'^a;  =  1. 
sin^a;  =  \. 
sin  a;  =  ^  \/2  = 
sin  J  a;  or  sin  22°  30^ 


cos  a;. 


_     ll-^\/2 


=  aV2-V2 


2 

=  0.3827. 
cos  A  a;  or  cos  22°  30' 


^ 


l  +  iV2 
2 


=  aV2+V2 


2 
=  0.9239. 


tan  \  x 


sm  *  a; 


A  a;     AV2-V2 


2i''_  f 


cosja;      1V2+V2 


2 


multiply  by 


^%+V2 
2-V2 
2-V2' 


cot  2  ^  = 


_     |(2-V2f 
\     4-2 

=  ^  V(2  -  V2)2  X  V2 
=  (1-JV2)XV2 

=  \/2- 1  =  0.4142. 

cos  I  a; 
sin  ^x 

^V2+V2 
iV2-\/2 


I2+V2 


2-V2 
=  \/2 +  1  =  2.4142. 


25.   Given  sin  30°  =  0.5 ;  find  the 
functions  of  15°. 

sin  30°  =  0.5  =  -• 
2 


cos  30° 


=V-^V! 


=  iV3. 


sin  ^  a;  =  ^- 


cosa; 


sin  15°  = 


l-iV^ 


=  jV2-V3  =  0.25885. 

cos  15°  =  -v — —^ 

\       2 

=  iV2  +  V3  =  0.96592. 


TEACHERS     EDITION. 


75 


tan  15°  = 


^1  +  ^ 


2V3 


-4 


2- V3 
2  + V3 


2-V3^2-V3 

2  +  V3     2  +  V3 


(2-V3)2 


4-3 
=  2- \/3  =  0.26799. 


cot  15° 


-V; 


+  ^V3 


iV3 
=  2  + V3  =  3.7321. 

26.  Prove  that 
^^^13o^sjn33^+sin3_°. 

cos  33°  + cos  3° 

Let        X  =  18°, 

y  =  15°. 

Then 

(1)  2 sin X  cosy 

=  sin  (a;+y)  +  sin  {x—y). 

(2)  2  cos  a;  cosy 

=  cos  (x+y)  +  cos  {x—y). 
Divide  (1)  by  (2). 

tan  a;  =  sm  (a;+y)  + sin  (a;-y) 

cos  {x+y)  +  cos  {x—y) 

Substitute  values  of  x  and  y, 

^^^^go_sjn33^J:_sin3f_ 
cos  33°  + cos  3° 

27.  Prove  the  formula 

•    o  2  tan  x 

sin  2  a;  = 

1  +  tan^a; 

sin  2  a;  =  2  sin  x  cos  x. 


2  tan  a;  = 


z  sma; 
cos  a; 


1  +  tan^j;  =  1  + 


cos^j; 


_  cos^a;  +  sin^a; 
cos^a; 

But  cos^a;  +  sin^a;  =  1. 

.-.  1  +  tan^a;  =  -1-. 
cos^a; 

2sina^cosaj  =  2siT^^cos^^ 
cos  a?  1 

•.  2  sin  X  cos  a;  =  2  sin  a:  cos  x. 


28.   Prove  the  formula 
1  —  tan^a; 


cos  2  a;  = 


1  +  tannic 


1 


cos  2  a;  = 


cos^a? 


1  + 


sin-'a; 


cos'^aj 
cos  2  a;  =  cos^a;  —  sin^a;. 


1- 


cos^a;  —  sin^a;  = 


ein^a? 

cos'-^a; 


1  + 


sin^B 
cos^a; 


_  cos%  —  sin^a; 
cos^a;  +  sin^a; 

_  cos^a;  —  sin^a; 

~        i 

=  cos^a;  —  sin^aj. 

29.   Prove  the  formula 
sin  a; 


tan  i  a;  = 


tan  2  a;  = 


1  +  cos  X 


vT^ 


cos  a? 


vr 


+  cos  a; 


76 


TRIGONOMETRY. 


sin  X 

Vi- 

-  CO'&^X 

1  +  cos  X 

1  + 

COS  X 

Vl  —  COS.T  _ 

Vi- 

-  cos^aj 

1  +  cos  X 


Vi  +  cos  X 
1  —  COS  X       1  —  cos^a; 


1  ±  2  sin  \ X  cos  ^x  =  l  ±  sin  x. 
±  2  sin  J  a;  cos  J  a;  =  ±  sin  a;. 

2  sin  ^  X  cos  J  x 


1  4-  cos  X      (1  +  cos  «)^ 
_  1  —  cos  a; 
1  +  cos  X 

30.  Prove  the  formula 


cot  J  a;  = 


sm  X 


1  —  cos  X 
sin  a;  =  Vl  —  cos'^a;. 

^  1  —  cos  X 

By  substituting, 


V 


1  +  cos  X       Vl 


cos^'a; 


1  —  cos  X  1  —  cos  X 
1  +  cos  X  _  1  —  cos^a; 
1  — cosa;     (1  — cosa;)^ 

_  1  +  cos  X 
1  —  cos  X 

31.   Prove  the  formula 

sin  I  a;  db  cos  J  a;  =  Vl  db  sin  x. 

By  squaring, 
sin^  J  a?  ±  2  sin  J  a;  cos  J  a;  +  cos^  |  x 
=  1  ±  sin  X. 


But  sin  h  X 


=V^ 


COS  a; 


and        COS  lx  =  -V 


/I  +  COS  a; 


—  cos'^.r 


±  sm  X 


=  ±A' 


/L 


cos  zx 


•.±2-^. 


1  —  cos'-^a; 


=  .^| 


1  —  coa  2  a; 


1  —  cos^x  = 


1  —  cos  2  aj 


Substitute   values  of  sin^a;  and 
1 


cos  f  x. 
1— cosx 


±2sin^a;cos|a;  + 
=  1  ±  sin  X. 


1  -f  cos  a; 


/-^ ^     ^h  —cos  2 a; 

Vl  —  cos''a;  =  •\l 

.'.  sin  a;  =  sin  a;. 

32.  Prove  the  formula 

t^M±l^M  =  ±tana;tan2/. 
cot  a;  ±  cot  y 

tan  a;  ±  tan  y 

=  ±  tan  X  cot  a;  tan  y 

+  cot  y  tan  y  tan  a; 

But  tan  a;  cot  a;  =  1 , 

and      tan3/coty  =  l. 

.-.  tan  X  ±  tan  y  ==  tan  aj  ±  tan  y. 

33.  Prove  the  formula 
1  —  tan  X 


tan  (45°  -  a;)  = 
tan  (45°  -  a;)  = 


1  +  tan  X 
sin  (45°  —  x) 
cos  (45°  —  x) 
sin  (45°  -  a;) 

=  sin  45°  cos  x  —  cos  45°  sin  x 


=  J  V2  cos  X  —  A  V2  si 


sm  X 


=  I V2  (cos  X  —  sin  a;), 
cos  (45°  —  a;) 

=  cos  45°  cos  X  +  sin  45°  sin  x 


TEACHERS     EDITION. 


=  ^  \/2  COS  a;  +  J  \/2  sin  a; 

=  2"\/2  (cos  tc  +  sin  x). 

4-      / 1  Ko        \      cos  X  —  sin  a? 
tan  (45°  —  x)  = -. — • 

cos  X  +  sin  X 

Dividing  numerator  and  denomi- 
nator by  cos  X, 

1  —  tan  X 


tan  (45°  -  a;)  = 


1  +  tan  X 


34.  If  A,  B,  C  are  the  angles  of 
a  triangle,  prove  that 

sin  A  +  B\n  B  -{■  sin  C 

=  4  cos  J  J.  cos  ^  ^  cos  I  C. 

Bin  ^  +  sin  j5  +  sin  C 

=  sin^+sin  ^+sin  [180°-(^+5)] 
=  sin  ^  +  sin  B  +  sin  {A  +  B) 
=  2  sin  ^  (^  +  B)  cos  1{A--B) 

+  2sin  H^  +  B)  cos  ^-  {A  +  5) 
=  2sini(^  +  B)  [cos  ^(^  -  B) 
+  cos  1(^  +  5)]. 

ButK^  +  -S)  =  ^^ 

Then,  by  [22], 
cos  A^-{-  cos  J5^ 

=  2  cos  ^  -4  cos  I  5. 
.-.  ==  4  sin  I  (J.  +  B)  cos  |  J.  cos  J  ^. 
But  cos  J  C=  cos  [90°  -1{A  +  B)] 

=  ^mi{A  +  B). 
.'.  sin  A  +  smB  +  sin  0 

=  4  cos  ^  -4  cos  ^  ^  cos  J  C. 

35.  If  ^,  5,  C  are  the  angles  of 
a  triangle,  prove  that 

cos  A  +  cos  B  +  cos  C 

=  1  +  4  sin  -J  ^  sin  J  ^  sin  |  C. 


cos  (7=  cos  [180°- (^  +  ^)] 

=  —  cos  (A  +  B). 
.'.  cos  A  +  cos  B  +  cos  C 

=  cos  J.  +  COS  B  —  cos  {A  +  5). 
By  [22], 

=  2  cos  J  (^  +  B)  cos  J  (^  -  B) 

—  cos  (J.  +  B). 

By  [17], 

=  2  cosK^  +  B)  cos  H^  -  -5) 

-  2  cos2  ^  (^  +  ^)  + 1 
=  [2cos^(^  +  ^)] 

X  [cos  i  Ia-B)-co&  ^{A+B)]-{-h 
By  [23], 

=  [2cosH^+^)] 

X  [2  sin  ^  J.  sin  ^  5]  +  1 
=  (2  sin  J  C)  (2  sin  ^  ^  sin  ^  5)  + 1 
=  1  +  4  sin  ^  ^  sin  J  .B  sin  ^  C. 

36.    If  A,  B,  C  are  the  angles  of 
a  triangle,  prove  that 

tan  A  +  tan  B  +  tan  C 

=  tan  ^  X  tan  B  X  tan  C. 

Since  ^  +  5  +  (7=  180°, 

C=1S0°-{A+  B). 

.-.  tan  C=  tan  [180°  -(^1  +  B)] 

=  -  tan  {A  +  ^). 

Again, 

tan  A  +  tan  5 

=  tan  {A  +  B){1  —  tan  A  tan  P) 

=  tan  {A  +  B) 

—  tan  {A  +  ^)  tan  ^  tan  B, 

.•.  tan  J.  +  tan  B  +  tan  Q 

=  tan  (^  +  ^)  -  tan  {A  +  ^) 

—  tan  {A  +  5)  tan  J.  tan  B 

=  -  tan  {A  +  5)  tan  ^  tan  B 

=  tan  j4  tan  B  tan  C. 


78 


TRIGONOMETRY. 


37.  If  A,  B,  C  are  the  angles  of 
a  triangle,  prove  tliat 

cot  ^  J.  +  cot  J  ^  +  cot  J  C 

=  cot  J  J.  X  cot  J  i?  X  cot  f  C. 

Since^^  + J5  +  iC=90°, 

.■.cotiC=ta,n^{A  +  B), 
and  cot  J  ^  =  tan  H^  +  Q^ 
and  cot  J  J.  =  tan  J  (^  +  C). 
.'.  cot  J  J.  +  cot  f  ^  +  cot  J  C 

=  tan  J  ( J.  +  5)  +  tan  i{A+  C) 

+  t&n}{B+C) 
=  tan  J  ( A  +  -g)  X  tan  J  (^  +  C) 
X  tan  K^  +  Q- 
By  substitution, 
cot  2"  -4.  +  cot  J  ^  +  cot  J  C 
=  cot^ilX  cot  ^  5  X  cot  J  C 

38.  Change  to  a  form  more  con- 
venient for  logarithmic  computation 
cot  X  +  tan  X. 

cot  X  +  tan  aj 
_  cos  X     sin  a; 
sin  X     cos  a; 
_  cos^a;  +  sin^g; 

sin  X  cos  X 
_  2  (cos^a;  +  sin^o;) 
2  sin  a;  cos  x 
2 
sin  2  a; 

39.  Change  to  a  form  more  con- 
venient for  logarithmic  computation 
cot  X  —  tan  X. 

cot  X  —  tan  X 
_  cos  K  _  sin  X 
sin  a;     cos  x 
_  cos%  —  sin% 
sin  X  cos  a; 


cos  2x 
sin  a;  cos  a; 
2  cos  2  a; 
2  sin  X  cos  a; 
2  cos  2  a; 
sin  2  a; 
2  cot  2  a;, 


[13] 


[12] 


40.  Change  to  a  form  more  con- 
venient for  logarithmic  computation 
cot  X  +  tan  y. 

cos  a;    J. sin  y 

cosy 


cot  a;  =  ^^^^^^,    tan?/=^^^^-^-       [2] 


sin  X 
Adding, 

_  cos  X  cos  y  +  sin  x  sin  y 
sin  a;  cosy 

Substitute  for  cos  x  cos  ?/-l-sin  x  sin  y 
its  equal  cos  {x  —  y),  [9] 

_  cos  (x  —  y) 
sin  X  cos  y 

41.  Change  to  a  form  more  con- 
venient for  logarithmic  computation 
cot  X  —  tan  y. 


,             sm  v 

tany  = ^• 

cosy 

,         cos  a; 
cot  X  = 

sm  X 

cot  a;  —  tan  y 

cos  a; 

sin  y 

sin  a; 

cosy 

cos  X  cos  y  —  sin  a;  sin  y 

sin  a;  cos  y 

_  cos  (a;  +  y) 

sm  a;  cos  y 

42.  Change  to  a  form  more  con- 
venient for  logarithmic  computation 
1  —  cos  2  a; 


1  -h  cos  2  a; 


TEACHERS     EDITION. 


79 


1  —  cos  2x 
1  +  cos  2x 


1 

—  cos 

2 

X 

2 

1 

+  cos 

2 

X 

2 
sin^a; 

cos-'a; 
=  tan^a;. 

43.  Change  to  a  form  more  con- 
venient for  logarithmic  computation 
1  +  tan  X  tan  y. 

1  +  tan  X  tan  y 

^      sin  X      sin  v 

=  1  + X 

cos  X      cos  2/ 

cos  X  cos  y  +  sin  x  sin  3/ 
cos  X  cos  3/ 

_cos(a;  — y) 
cos  a;  cosy 

44.  Change  to  a  form  more  con- 
venient for  logarithmic  computation 
1  —  tan  X  tan  y. 

1  —  tan  X  tan  y 
_  ^      sin  X  sin  3/ 
cos  X  COS  y 

_  cos  X  COS  2/  —  sin  x  sin  3/ 
COS  X  cos  2/ 
cos  (a;  -f  y) 
COS  a;  cos  y 

45.  Change  to  a  form  more  con- 
venient for  logarithmic  computation 
cot  a;  cot  2/ +  1. 

cot  X  cot  2/  +  1 


cos  a;     cos  y 
sm  X     sin  2/ 


+  1 


By  [9] 


_  COS  X  cos  2/  +  sin  a;  sin  2/ 
sin  X  sin  3/ 

_  cos  {x  —  y) 
sin  X  sin  2/ 


46.  Change  to  a  form  more  con- 
venient for  logarithmic  computation 
cot  xcoiy  —  1. 

cot  X  cot  3/  —  1 


cos  a;  cos  y 


-1 


sm  a;  sm  y 

_  cos  a;  cos  3/  —  sin  a;  sin  3/ 

sin  a;  sin  3/  ^ 

_  cos  [x  +  3/) 
sin  a;  sm  y 

47.   Change  to  a  form  more  con- 
venient for  logarithmic  computation 
tan  X  +  tan  y 
cot  a;  +  cot  y 

tana;  -f  tang/ 
cot  a;  -h  cot  y 

sin  a;  sin  3/ 

cos  X  cos  2/ 

cos  a;  cos  2/ 

sin  a;  sin  2/ 

sin  a;  cos  y  +  co&x  sin  3/ 
cos  X  cos  2/ 


sm  a;  cos  y  +  cos  a;  sm  y 
sin  a;  sin  y 

sin  a;  sin  2/ 


cos  X  cos  2/ 
tan  X  tan  y. 


80 


TBIGONOMETEY. 


Exercise  XII.     Page  53. 


1.  What  do  the  formulas  of  §  36 
become  when  one  of  the  angles  is  a 
right  angle  ? 

B 


If  angle  C  is  a  right  angle, 

a     sin  A       ■      . 

-  = =  sm  A 

c      sin  C 

c  _  sin  C  _     1 
b     sin  B     sin  B 


a  _  sin  A 
b     sin  B 

a  c 


=  tan  A  ; 


sin  A     sin  C 

b     ^     c 
sin  B     sin  C 


=  c; 


=  c. 


2.  Prove  by  means  of  the  Law  of 
_Sines  that  the  bisector  of  an  angle 
of  a  triangle  divides  the  opposite 
side  into  parts  proportional  to  the 
adjacent  sides. 

Let  CD  bisect  angle  C. 


Then 

and 


AD  _sm^O 
CD       sin  A ' 

DB  ^  sin  ^  (7 
CD       sin  B ' 


By  division, 

AD  _  sin  B 
DB     sin  A 

But        ?iB4  =  ^. 
sin  A     a 

"  BD     a 

3.  What  does  Formula  [26]  be- 
come when  ^  =  90°  ?  when  ^  =  0°  ? 
when  ^-180°?  What  does  the 
triangle  become  in  each  of  these 
cases  ? 

Formula  [26]  is 

a^  =  b^  +  c^  —  2bc  cos  A. 

When  A  =  90°,  cos  A  =  0°. 

.-.  a^  =  b'^  +  c\ 

When  A  =  0°,  cos  A  =  l. 

.-.  a2  =  62  +  c2-26c. 

When  A  =  180°,  cos  ^  =  -  1. 
.•.a2  =  62  +  c2  +  2Jc. 

B 


A 

a  =  BC. 

b  =  AC 

B 


a  =  Ba 
b^AC 


0 

c=AB. 
a  =  b  —  c. 

0 

c=BA. 

a  =  b  +  c. 


4.  Prove  that  whether  the  angle 
B  is  acute  or  obtuse  c  =  a  cos  B 
+  b  cos  A.  What  are  the  two 
symmetrical  formulas  obtained   by 


TEACHERS     EDITION. 


81 


changing  the  letters?     What  does 
the  formula  become  when  B  =  90°  ? 


A 

Fig.  2. 

D 

Case  I 
(Fig.  1). 

(1) 

.   When  angle  B 
a 

is 

acute 

h 

,'.  DB  =  a  cos  B, 
and  AD  =  h  cos  A. 

Add,  Z)5+^i)  =  acos5  +  6cos^. 
But    Z)5+^i)  =  c. 

.".  c  =  a  cos  5  +  6  cos  A. 

Case  II.  When  angle  B  is  obtuse 
(Fig.  2), 


^D 


=  cos  ^. 


:?:^  =  cos(180°-.B) 
=  —  cos  ^. 


.-.  AD  =  6cos^, 

and  5Z)  =  —  a  cos  B. 

Subtract,  observing  that  the  sign 
of  cos  B  is  minus. 

AD  —  BD  =  6  cos  J.  +  a  cos  B. 

But  AD-BD  =  c. 

,'.c  =  acosB  +  b  cos  A 

The  symmetrical  formulas  are 

b  =  a  cos  C+  ccos^, 

a  =  6  cos  C  +  c  cos  -S. 

When  B  =  90°. 

(3)         cos  ^  =  -• 
6 

.'.  c  =  6  cos  -4. 

5.  From  the  three  following  equa- 
tions (found  in  the  last  exercise) 
prove  the  theorem  of  ^  37  : 

c  =  a  cos  5  +  6  cos  J., 

6  =  a  cos  C  +  c  cos  A^ 

a  =  6  cos  (7  +  c  cos  B. 

&  =  ac  cos  B  -{■  be  cos  A.  (1) 
52  =  a6  cos  C+bc  cos  J..  (2) 
a^  =  ab  cos  C  +  ac  cos  .8.         (3) 

Add  (2)  and  (3), 
a'  +  b^  =  2ab  cos  C+bc  cos  .4 
+  ac  cos  j5.  (4) 

Subtract  (4)  from  (1), 

c^  —  a^  —  b'^  =  —  2ab  cos  0. 

.•.c2  =  a2  +  62_2a6cosa      ^7 

6.  In  Formula  [27]  what  is  the 
maximum  value   of  ^{A  —  B)l  of 

a  —  b  _  tan  ^(A  —  B) 
a+b     tanJ(^  +  5) 


82 


THIGONOMETRY. 


The  limit  of  ^  -  5  is  180°. 
.'.  the  limit  of  the  maximum  value 
ofH^-^) 


180° 
2 


=  90°. 


The  limit  of  ^  +  5  is  180°. 

/.  the  limit  of  the  maximum  value 
of^{A  +  B) 

=  1^0!  =90° 
2 

7.  Find  the  form  to  which  Form- 
ula [27]  reduces,  and  describe  the 
nature  of  the  triangle  when 

(i.)  C=90°; 

(ii.)  A- B  =  90°,  and  B=0. 
a  —  b  ^  tan  ^{A  —  B) 
a  +  b      isin^{A  +  B) 

(i.)  When  C=  90°. 
A  +  B  =  90°. 

B=90°-A. 

Q-5^tann^-(Q0°-^)1 
a  +  b  tan  45° 

_^  tan  (^-45°) 

1 
=  tan  (^-45°). 


Since  C  is  a  right  angle,  the  tri- 
angle is  a  right  triangle. 

(ii.)  When  ^-5  =  90°,  and  5=a 

a  —  b  _  tan  ^{A  —  B) 
a  +  b     tan  ^  ( J.  +  5) 

A  +  B +  0^180°, 

or      A  +  2B      =  180° 

A-B        =90° 


,'. 

3  5=   90° 

B=   30°, 

C=   30°, 

and 

A  =  120°. 

a  —  b 

tan  45° 

a  +  b 

tan  75° 

tan  45° 

cot  15° 

1 

2+VS 

•'• 

a  +  b 

=  (a-6)(2  +  >/3). 

Since  A  ■- 
celes. 

=  B,  the  triangle  is  isos 

1.  Given 
a  =  500, 
A  =  10°  12^ 
B  =  46°  36' ; 


Exercise  XIII. 

find 
C=  123°  12^ 
b  =  2051.48, 
c  =  2362.61. 


a  =  500. 
A=   10°  12' 
B  =   46°  36' 


A  +  B=   56°  48' 
J.  C=  123°  la^ 


Page  55. 

log  a        =  2.69897 

colog  sin  A  =  0.75182 

log  sia  B  =  9.86128 

log  b        =  3.31207 
6  =  2051.48. 

log  a        =  2.69897 

colog  sin  A  =  0.75182 

log  sin  C  =  9.92260 

logc         =3.37339 
c  =  2362.61. 


TEACHERS     EDITION. 


83 


2.   Given  find 

a  =  795,  C=  55°  20^ 

J.  =  79°59^  6=567.688, 

B  =  44°  4F;  c  ==  663.986. 

a  =  795. 

^=  79°  59^ 

B  =  44°  4F 


A-\-B=  124°  40^ 
.-.  C=   55°  20'. 

log  rt  =  2.90037 
colog  sin  ^  =  0.00667 
log  sin  B  =  9.84707 
log  b  =  2.75411 
b  =  567.688. 

log  a        =  2.90037 

colog  sin  A  =  0.00667 

log  sin  C  =  9.91512 

log  c         =  2.82216 
c  =  663.986. 

3.  Given  find 

a  =  804,  C=35°4^ 

^  =  99°55^  6  =  577.31, 

^  =  45°  1' ;  c  =  468.93. 

a  =  804. 

A=   99°  55' 
^=   45°    V 


A  +  B  =  144°  56' 
.-.  C=   35°    4'. 

log  a  =  2.90526 
colog  sin  A  =  0.00654 
log  sin  B  =  9.84961 
log  6  =  2.76141 
6  =  577.31. 


log  a.        =  2.90526 

colog  sin  A  =  0.00654 

log  sin  Q  =  9.75931 

log  c         =  2.67111 

c  =  468.93. 

4.  Given  find 

a  =  820,  C=  25°  12', 

A  =  12°  49',  6  =  2276.63, 

5  =141°  59';         c  =  1573.89. 

a  =  820. 

A=   12°  49' 
^  =  141°  59' 

^  +  .5  =154°  48' 
.-,  C=   25°  12'. 

log  a        =  2.91381 

colog  sin  A  =  0.65398 

log  sin  B  =  9.78950 

log  6         =3.36729 
b  =  2276.63. 

log  a        =  2.91381 

colog  sin  A  =  0.65398 

log  sm  0  =  9.62918 

log  c         =  3.19697 
c  =  1573.89. 

5.  Giv^n  find 

c  =  1005,  C=  47°  14', 

A  =  78°  19',  a  =  1340.6, 

^  =  54°  27';  ,6  =  1113.8. 

c  =  1005. 

^=   78°  19' 

B  =   54°  27' 

^  +  ^  =  132°  46' 

.-.  C=   47°  14'. 


84 


TRIGONOMETKY. 


logc 

=  3.00217 

colog  sin 

C 

=  0.13423 

log  sin 

A 

=  9.99091 

log  a 

=  3.12731 

a 

=  1340.6. 

log  c         =  3.00217 

colog  sin  C  =  0.13423 

log  sin  B  =  9.91042 

log  b         =  3.04682 
b  =1113.8. 


6.   Given 
b  =  13.57, 
5=13°57^ 
(7=57°  13^ 


find 
^  =  108°50^ 
a  =  53.276, 
c  =  47.324. 


6  =  13.57. 

£^    13°  57^ 
C=   57°  13' 


£  +  0=    71°  10' 
.-.  A  =  108°  50'. 

log  b         =  1.13258 

colog  sin  B  =  0.61785 

log  sin  A  =  9.97610 

log  a         =  1.72653 
a  =  53.276. 

log  a        =  1.72653 

colog  sin  A  =  0.02390 

log  sin  C  =  9.92465 

log  c         =  1.67508 
c  =  47.324. 


7.  Given 
a  =  6412, 
A  =  70°  55', 
C  =  52°  9'  ; 


find 

B  =  56°  56', 

6  =  5685.9, 

c  =  5357.5. 


a  =  6412. 

A=    70°  55' 

C=    52°    9' 

A+ 0=123°    4' 

.-.  B  =    56°  56'. 

log  a        =  3.80699 

log  sin  B  =  9.92326 

colog  sin  A  =  0.02455 

log  6         =  3.75480 
b  =  5685.9. 

log  a        =  3.80699 

log  sin  C  =  9.89742 

colog  sin  A  =  0.02455 

log  c         =  3.72896 
c  =  5357.5. 


8.  Given 

b  =  999, 
A  =  37°  58', 

C=  65°  2' ; 


find 
B  =  77°, 
a  =  630.77, 
c  =  929.48. 


6  =  999. 

A=    37°  58' 
C=    65°    2' 


.4  +  C=103° 
.•.^=   77°. 

log  b         =  2.99957 

colog  sin  5=0.01128 

log  sin  A  =  9.78902 


log  a 

=  2.79987 

a 

=  630.77. 

log  6 

=  2.99957 

colog  sin 

B 

=  0.01128 

log  sin 

C 

=  9.95739 

logc 

=  2.96824 

c 

=  929.48. 

TEACHERS     EDITION. 


85 


9.  In  order  to  determine  the  dis- 
tance of  a  hostile  fort  A  from  a  place 
B,  a  line  BC  and  the  angles  ABC 
and  BCA  were  measured,  and  found 
to  be  322.55  yards,  60°  3V,  and 
56°  10^,  respectively.  Find  the  dis- 
tance AB. 

a  =  322.55. 


B  = 

60°  34^ 

C= 

56°  10^ 

B+C= 

116°  44^ 

/. 

A  = 

63°  16^ 

log  a 

2.50860 

colog  sin 

A  = 

0.04910 

log  sin 

C  = 

9.91942 

logc 

2.47712 

c 

300. 

10.  In  making  a  survey  by  tri- 
angulation,  the  angles  B  and  Cof  a 
triangle  ABC  were  found  to  be 
50°  30^  and  122°  9^  respectively, 
and  the  length  BC  is  known  to  be 
9  miles.     Find  AB  and  AC. 

(7=122°    9^ 
B=    50°  30^ 


B+C= 

-.  172°  39^ 

.'.  A  = 

=     7°  21^ 

log  BC    = 

0.95424 

colog  sin  A  = 

0.89303 

log  sin  B  = 

9.88741 

log  b 

1.73468 

b  =  AC    = 

54.285. 

log  BC    = 

0.95424 

colog  sin  A  = 

0.89303 

log  sin  C  = 

9.92771 

logc 
c^AB 


1.77498 
59.564. 


11.  Two  observers  5  miles  apart 
on  a  plain,  and  facing  each  other, 
find  that  the  angles  of  elevation  of 
a  balloon  in  the  same  vertical  plane 
with  themselves  are  55°  and  58°, 
respectively.  Find  the  distance 
from  the  balloon  to  each  observer, 
and  also  the  height  of  the  balloon 
above  the  plain. 

B=   58° 
A=    55° 


A  + 

B 

=  113° 

C 

=   67°. 

logc 

=  0.69897 

colog  sin 

c 

=  0.03597 

log  sin 

A 

=  9.91336 

log  a 

=  0.64830 

a  =  BC 

=  4.4494. 

logc 

=  0.69897 

colog  sin 

C 

=  0.03597 

log  sin 

B 

=  9.92842 

log  6 

=  0.66336 

h^AC 

=  4.6064. 

)  find  h. 

h 
a 

=  sin  B. 

.• 

.h 

=  a  sin  B. 

log  a 

=  0.64830 

log  sin 

B 

=  9.92842 

log  A 

=  0.57672 

h 

=  3.7733. 

12.  In  a  parallelogram,  given  a 
diagonal  d  and  the  angles  x  and  y 
which  this  diagonal  makes  with  the 
sides.      Find   the   sides.     Compute 


86 


TRIGONOMETRY. 


the  results   when   cZ  =  11.237,   x 
Id"  V,  and  y  =  42°  5V. 

d=  11.237. 

x=    19°    V 
2/=   42°  54^ 

x  +  y=    61°  55^ 
.-.  z  =  118°    5^ 

\ogd        =  1.05065 

colog  sin  z   =  0.05440 

log  sin  X  =  9.51301 


log  a 

=  0.61806 

a 

=  4.1501. 

logd 

=  1.05065 

colog  sin 

2 

=  0.05440 

log  sin 

y 

-  9.84297 

logc 

=  0.93802 

c 

=  8.67. 

13.  A  lighthouse  was  observed 
from  a  ship  to  bear  N.  34°  E. ;  after 
sailing  due  south  3  miles,  it  bore  N. 
23°  E.  Find  the  distance  from  the 
lighthouse  to  the  ship  in  both  posi- 
tions. 

c  =  3. 


5  =  (180^ 


A=   23° 

'-34°)  =  146° 

A  +  B  =  169° 

.\C=   11°. 


log  c  =  0.47712 
colog  sin  C  =  0.71940 
log  sin  A  =  9.59188 
log  a  =  0.78840 
a  =  6.1433. 


log  c         =  0.47712 

colog  sin  Q  =  0.71940 

log  sin  B  =  9.74756 

log  h        =  0.94408 
h  =  8.7918. 

14,  In  a  trapezoid,  given  the 
parallel  sides  a  and  6,  and  the  an- 
gles X  and  y  at  the  ends  of  one  of 
the  parallel  sides.  Find  the  non- 
parallel  sides.  Compute  the  results 
when  a  =  15,  6  =  7,  a;  =  70°,  y  =  40°. 

Given  parallel  sides, 

^5  =7  and        DC=15; 
also,  ADG=  40°  and  BCD  =10°; 
required  AD  and  BC. 

Draw  AE  II  BC; 
then     AB  =  EC    (lis  comp.  bet.  lis), 
and      DE=DC-AB 
=  15  -  7  =  8. 
Also  AED  =  BCD=10°{ext.  int.  A). 
Now 

i)JL.£;=180°-(40°  +  70°) 
=  70°. 
But  since 

AED  =  DAE  =70°, 

the  A  is  isosceles,  and  side 
DA=DE=8. 

'Now  AE=  BC,  and  we  are  to  find 
BC. 

AE_  sin  ADE 

DE     sin  DAE 
log  DE  =  0.90309 

log  sin  ADE  =9.80807 
colog  sin  DAE  =  0.02701 

log  AE      -     =  0.73817 
AE=BC        =5.4723. 


TEACHERS     EDITION. 


87 


15.  Given  h  =  7.07107,  A  =  30°, 
C=  105° ;  find  a  and  c  without 
using  logarithms. 

Let  p  and  q  denote  the  segments 
of  c  made  by  the  ±  dropped  from  C. 

5  =  45°. 

sin  A=  — 

2 


sin 


5  =  iV2. 


a_ 
b~ 

a  = 


i 


iV2 
_b_ 
V2 
7.07107 


=  5. 


b 

P 


1.41421 

cos  ^  =  |\/3  =  0.86602. 

=  bx  0.86602 

=  7.07107  X  0.86602 

=  6.12369. 


?  =  sin  5  =  JV2  =  0.70711. 

q  =  ax  0.70711 

=  5  X  0.70711  =  3.53555. 
c=p  +  q 

=  6.12369  +  3.53555 

=  9.65924. 

16.  Given  c  =  9.562,  A  =  45°, 
B  =  60° ;  find  a  and  b  without 
using  logarithms. 

C=  75°. 

e  sin  A 


a  = 


=  ^V2xiV3  +  ^V2xi 
9.562  x*V2 


b  = 


(v/6+V2) 
_  19.124  xV2 
V6+V2 

_(19.124xV2)(\/6-V2) 

6-2 
=  9.562  (VS-l) 
=  6.999  =  7. 

a  sin  5      7  X  iVS 


sin  J. 
7V3     7\/6 


f  V.4J 


a  = 


sin  C 


sin  a=  sin  (45° +  30°) 
=  sin  45°  cos  30° 

+  cos45°sin30°. 


V2         2 
=  3.5  V6  =  8.573. 

17.  The  base  of  a  triangle  is  600 
feet,  and  the  angles  at  the  base  are 
30°  and  120°.  Find  the  other  sides 
and  the  altitude  without  using  log- 
arithms. 

AB  =  600. 
A  =  30°. 
5=120°. 
.-.  C=  30°. 
A=C. 

a  =  c  =  600  feet. 
7  _  a  sin  5 
sin  A 
_  600  X  sin  (180°- 60°) 
sin  30° 

^600x|a/3 
1 

=  600x1.732051 
=  1039.2. 
h  =  b  sin  A  ==  1039.2  X  J 
=  519.6  feet. 


TEIGONOMETRY. 


18.  Two  angles  of  a  triangle  are, 
the  one  20°,  the  other  40°.  Find 
the  ratio  of  the  opposite  sides  with- 
out using  logarithms. 

Let  X  =  20°, 

37  =  40'', 

and  a  and  h  be  opposite  sides. 

sin  X  _a 


Then 


sin  y 
nat  sin  a;  =  0.3420. 
nat  sin  y  =  0.6428. 
.-.  a  :  6  :  :  3420  :  6428. 
:  :    855  :  1607. 

19.  The  angles  of  a  triangle  are 
as  5  :  10  :  21,  and  the  side  opposite 
the  smallest  angle  is  equal  to  3. 
Find  the  other  sides  without  using 
logarithms. 

Since  the  angles  A,  B,  C  are  as 
5:10:21, 

^  =  ^6  of  180°  =  25°. 

^=10  of  180°  =  50°. 

C=|iof  180°  =  105°. 


a  sin  5      3  X  0.766 


sin  J. 

=  5.43775. 

a  sin  C 
sin  A 

=  6.857. 


c  = 


0.4226 


3  X  0.9659 
0.4226 


20.  Given  one  side  of  a  triangle 
equal  to  27,  the  adjacent  angles 
equal  each  to  30°.  Find  the  radius 
of  the  circumscribed  circle  without 
using  logarithms. 

2E 


sin^ 
sin  A  =  sin  120° 

=  sin  (180°  -  60°) 
=  sin  60°. 
sin  60°  =  |\/3. 

27       _54  _  54xV3 
V3  3 


•.2i2  = 


^V3 
=  18  Vs. 
i2=9V3  =  15.588. 


Exercise  XIV.     Page  59. 


1.    Determine  the  number  of  solu- 
tions in  each  of  the  following  cases  : 

(i.)  a  =  80,  6  =  100,  A  =  30°. 

a  =  80  <  6  =  100, 
but  >  5  sin  ^  =  100  X  |, 

and  A  <  90°. 

.*.  two  solutions. 


(ii.)  a  =  50,  ft  =  100,  ^  =  30°. 

a  =  50  =  6  sin  il  =  100  X  h 
.'.  one  solution. 


(iii.)  a  =  40,  6  =  100,  ^  =  30°. 

a  =  40<  bsmA^lOOxh 
and  A  <  90°. 

.'.  no  solution. 

(iv.)  a  =  13.4,  6  =  11.46,  ^=77° 20'. 
a  =  13.4  >  5=11.46. 
.*.  one  solution. 

(v.)  a  =  70,  5  =  75,  A  =  60°. 

a  =  70  <  &  =  75, 
but  >6sin^  =  75xiV3, 


TEACHERS     EDITION. 


89 


and  ^  =  60°  <  90°. 
.*.  two  solutions. 

(vi.)  a  =  134.16,  b  =  84.54, 

B  =  52°  9^  11^^. 
b<a,      B<  90°, 
nat  sin  B  =  0.7897. 
.*.  &  <  a  sin  B. 

84.54  <  134.16  X  0.7897. 
.*.  no  solution. 


2.   Given 
a  =  840, 
b  =  485, 
^  =  21°31^ 


find 
B=   12°13^34'^ 
C=146°15'26^^ 
c  =  1272.15. 


Here  a^  b,  and  log  sin  B  <.0. 

.'.  one  solution, 
colog  a         =  7.07572  -  10 
log  b         =  2.68574 
log  sin  A  -  9.56440 

log  sin  B  =  9.32586 
B  =    12°  13^  34^^ 

.-.  0  =  146°  15^  26^^ 

log  a        -  2.92428 

log  sin  C  =  9.74466 

colog  sin  A  =  0.43560 

log  c         =  3.10454 
c  =  1272.15. 


3.   Given 
a  =  9.399, 
6  =  9.197, 
^  =  120°  35^ 

colog  a 
log  h 


find 
B  =  57°  23^  40^^ 
C=    2°    1^20^^ 
c  =  0.38525. 

=  9.02692  -  10 

=  0.96365 


log  sin  A  =  9.93495 
log  sin  B  =  9.92552 
B 


=  57°  23^  40^^ 
2°    F20^^ 


log  a        =  0.97308 

log  sin  (7  =  8.54761 

colog  sin  A  =  0.06505 

logc 
c 


=  9.58574  - 10 
=  0.38525. 


4.   Given 
a  =  91.06, 
b  =  77.04, 
A  =  51°  9^  6^^ ; 

colog  a 
log  6 


find 
B  =  41° 13^ 
C  =  87°  37^  54^^ 
c  =  116.82. 

=  8.04067  - 10 

=  1.88672 


log  sin  A  =  9.89143 


log  sin  B  =  9.81882 
B  =  41°  13^ 

.-.  0  =  87°  37^  54^^ 

log  a  =1.95933 
log  sin  C  =  9.99963 
colog  sin  A  =  0.10857 
log  c  =  2.06753 
c  =  116.82. 

5.   Given  find 

a  =  55.55,  A  =  54°  3F  13^^, 

6  =  66.66,  (7  =  47°  44^    7^^, 

B  =  77°  W  4.0'' ;   c  =  50.481. 

Here  b^  a,  and  log  sin  A  <  0. 
.•.  one  solution. 

log  a        =  1.74468 
log  sin  B  =  9.98999 
colog  b         =8.17613-10 
log  sin  A  =  9.91080 
A  =  54°  31^  13^^. 

.\G  =47°  44^    7''. 


90 


TRIGONOMETRY. 


log  a        =  1.74468 

log  sin  C  =  9.86925 

colog  sin  A  =  0.08920 

log  c         =  1.70313 
c  =  50.481. 

6.   Given 
a  =  309,  6  =  360,  ^  =  21°14'25^^ 
find         B=    24°57^54^^ 
^^=155°    2'    6'^, 
C=133°47MF^ 
C'=     3°  43^  29'^ 
c  =  615.67. 
c^=  55.41. 

log  b         =  2.55630 
log  sin  A  =  9.55904 
colog  a         =  7.51004  -  10 

log  sin  B  =  9.62538 
B  =   24°  57^  54^^. 

.-.  C  =  133°  47^  41^^. 

log  a  =  2.48996 
log  sin  C  =  9.85843 
colog  sin  A  =  0.44096 
log  c  =  2.78935 
c  =  615.67. 

Second  Solution. 

^/=  180°  -  B 
=  155°  2^  6'\ 

C'=B-A 

\  =  3°  43^  29^^ 

log  a        =  2.48996 

log  sin  C^  =  8.81267 

colog  sin  A  =  0.44096 

log  c^        =  1.74359 
c^  =  55.41. 


7.   Given  a  =  8.716, 
h  =  9.787, 
^=   38°  14:^12^^; 
find  .5  =    44°    V  28^^ 

5^=  135°  58^  32^^ 
e=    97°44^20^^ 
(7=     5°  47^  16^^ 
c  =  13.954, 
c^=  1.4203. 

Here  a  <ih,  and  log  sin  5  <  0. 
.•.  two  solutions. 


colog 

a 

9.05968  - 

-10 

log 

b 

0.99065 

log 

sin 
sin 

^  = 
B  = 

9.79163 

log 

9.84196 

.5  = 

44°    V 

28^^ 

5^= 

135°  58^ 

32^^ 

G=   97°44'20^^ 
C'=     5°  47^  16^^ 

log  a  =  0.94032 
log  sin  C  =  9.99602 
colog  sin  A  =  0.20837 
log  c  =  1.14471 
c  =  13.954. 

log  a  =  0.94032 
log  sin  (7=9.00365 
colog  sin  A  =  0.20837 
log  c^  =  0.15234 
c^  =1.4203. 


8.   Given  find 

a  =  4.4,  B  =  90°, 

6  =  5.21,  (7=32°22M3^^ 

^  =  57°  37^  17^^'  c=2.79. 


TEACHERS     EDITION". 


91 


log  sin  A  =    9.92661 
log  b         =    0.71684 
colog  a         =    9.35655  -  10 

log  sin  B  =  10.00000 
£  =  90°. 

.-.  C  =  32°  22'  43'^ 

log  b  =  0.71684 
log  cos  ^  =  9.72877 
log  c  =  0.44561 
c  =2.791. 

9.  Given   a  =  34, 
6  =  22, 
B  =  30^  20' ; 
find  A  =    51°  18'  27", 

^/=  128°  41' 33", 
(7=    98°  21' 33", 
(7'=    20°  58' 27", 
c  =  43.098, 
c'=  15.593. 

Here  b  <^a,  but  >»  a  sin  J5,  and 
^  <  90°. 

.•.  two  solutions 

log  a        =  1.53148 

log  sin  B  =  9.70332 
colog  b         =  8.65758  -  10 

log  sin  A  =  9.89238 

A  =    51°  18'  27". 

A^=  128°  41'  33". 
.-.  C=  98°  21°  33". 
.-.  C'=    20°  58'  27". 

log  a  =1.53148 
log  sin  (7=9.99536 
colog  sin  A  =  0.10762 
log  c  =  1.63446 
c  =  43.098. 


log  a 

=  1.53148 

log  sin 

C"= 

=  9.55382 

colog  sin 

J.= 

=  0.10762 

log  c' 

=  1.19292 

c' 

=  15.593. 

10.  Given 

b  = 

=  19, 

c  = 

=  18, 

C  = 

=  15°  49'  ; 

nd 

B  = 

=    16°  43' 

13", 

^'= 

=  163°  16' 

47", 

A  = 

= 147°  27' 

47", 

A'= 

=     0°  54' 

13", 

a  = 

=  35.52, 

a'= 

=  1.0415. 

log  b 

:  1.27875 

log  sin 

(7  = 

9.43546 

colog  c 

8.74473  - 

-10 

log  sin 

5  = 

9.45894 

.5 

16°  43' 

13". 

^' 

- 

=  163°  16'  47". 

A  =  147°  2 

7'  47". 

^'=  0^ 

'54' 

13". 

log  6 

1.27875 

colog  sin 

B  = 

0.54106 

log  sin 

A  = 

9.73065 

log  a 

1.55046 

a 

= 

35.519. 

log  b 

= 

1.27875 

colog  sin 

5'= 

0.54106 

log  sin 

J.'= 

8.19784 

log  a' 

= 

0.01765 

a' 

= 

1.0415. 

11.    Given  <x  =  75,    5  =  29,    5  = 

16°  15'  36" ;  find  the  difi'erence  be- 
tween the  areas  of  the  two  corre- 
sponding triangles. 


92 


TRIGONOMETEY. 


log  a        =  1.87506 
colog  6         =  8.53760  -  10 
log  sin -8  =  9.44715 

log  sin  A  =  9.85981 

A  =    46°  2V  A5'\ 
A'=  133°  36^  15^^ 

(7=  117°  20^  39^^ 
C'=    30°    8^    9'^. 

log  a        =  1.87506 

colog  sin  A  =  0.14019 

log  sin  0  =  9.94854 

log  c         =  1.96379 

log  a        =  1.87506 

colog  sin  A  =  0.14019 

log  sin  (7''=  9.70075 


log  c^ 

=  1.71600 

i^=JcA. 

A  =  6  sin  A. 

log  h 

=  1.46240 

log  sin 

^  =  9.85981 

logh 

=  1.32221 

logc 

=  1.96379 

log/i 

=  1.32221 

colog  2 

=  9.69897 

logi^ 

=  2.98497 

i?' 

=  965.98. 

log  c^ 

=  1.71600 

log  A 

=  1.32221 

colog  2 

=  9.69897 

logi^'' 

=  2.73718 

F' 

=  545.99. 

F-F' 

=  419.99. 

12.  Given  in  a  parallelogram  tlie 
side  a,  a  diagonal  d,  and  the  angle 
A  made  by  the  two  diagonals  ;  find 
the  other  diagonal. 

Special  case :  a  =  35,  d  =  63, 
A=     °  36^  30^^ 

a  =  35. 
JcZ=31.5. 
J.  =  21°  36^  30^^ 

colog  a         =  8.45593  -  10 
log  J  J      =1.49831 
log  sin  ^1  =  9.56615 
log  sin  B  =  9.52039 
B  =    19°  21^  20^^. 

C  =139°    2^10^^ 

log  a        =  1.54407 

log  sin  C  =  9.81663 

colog  sin  A  =  0.43385 

logid^    =1.79455 
id^  =62.3085. 

d^  =  124.617. 


1,    Given 
a  =  77.99, 
b  =  83.39, 

C  =  72°  15' 


Exercise  XV. 

find 
A  =  51°  15^ 
B  =  56°  30^ 
c  =  95.24. 


a +  5  =  161.38 
6  -  a  =  5.4 


Page  62. 

A  +  B  =  107°  45' 
i{A  +  B)=    53°  52' 30'' 

J  (^  -  ^)  =     2°  37'  30" 
A=   51°  15'. 
5=   56°  30'. 


TEACHERS     EDITION. 


93 


log  (5  -  a)  =  0.73239 

colog  (a  +  6)  =  7.79215  -  10 

log  tan  ^{A  +  B)  =  0.13675 

log  tan  UB-A)  =  8.66129 

i{B-A)  =  2°  37^30^^. 


log  6 

=  1.92111 

log  sin  C 

=  9.97882 

colog  sin  B 

=  0.07889 

log  G 

=  1.97882 

c 

=  95.24. 

2.   Given  find 

b  =  872.5,  B  =  60°  45^ 

c  =  632.7,  C  =  39°  15^ 

^  =  80°  ;  a  =  984.8. 

b-c  =  239.8 
b  +  c=  1505.2 
B+C=  100° 
^{B  +  C)=   50° 

i{B-C)=  10°  45^ 
^=  60°  45^ 
(7=    39°  15^ 

log  (b-c)  =  2.37985 

log  tan  J  (5  +  C)  =  0.07619 
colog  (6  +  c)  =  6.82240  -  10 


log  tan  ^{B  - 

-C) 

=  9.27844 

UB- 

-C) 

=  10°  45^ 

logb 

=  2.94077 

log  sin  A 

=  9.99335 

colog  sin  B 

=  0.05924 

log  a 

=  2.99336 

a 

=  984.83. 

3.   Given 

find 

a  =  17, 

A  = 

=  77°  12^  53^^ 

6  =  12, 

B  = 

=  43°  30^    r^, 

C  =  59°  17^; 

c  = 

- 14.987. 

a  +  6  =  29 
a  —  6  =  5 
.A +  ^  =  120°  43^ 
1{A  +  B)=    60°  21^  30^^ 

1{A-B)=  16°  51^23^^ 
A=  77°12^53^^ 
B=   43°  30^   7^^ 

log  (a  -  6)  =   0.69897 

colog  (a +  5)  =    8.53760-10 

log  tan  ^{A  +  B)  =  10.24486 
log  tan  J  (A  -  5)  =    9.48143 

1{A-B)  =  16°  51^  23^^ 


log  6 
log  sin  C 
colog  sin  B 

logc 
c 


1.07918 
9.93435 
0.16218 

1.17571 
14.987. 


4.  Given  find 

b=Vb,  .5  =  93°  28^  36^^ 

c=V3,  C  =  50°38^24^^ 

A  =  35°o3^  a  =1.313. 

V5  =  2.2361 
V3  =  1.7321 
6 +  c  =  3.9681 
b-c  =  0.5040 
^  +  (7=144°   7^ 
1  (^  +  (7)  =    72°   3^  30^^ 
|(^-(7)=    21° 25^   &^ 
B=   93°28^36^^ 
C=    50°38^24^^ 

log(5-c)  =    9.70243-10 

colog  (b  +  c)  =    9.40142  -10 

log  tan  i{B  +  C)  =  10.48973 

logtan  J(^-C)=    9.59358 

J(J5-C)  =  21°25^6^^ 


94 


TRIGONOMETRY. 


logc 

=  0.23856 

log  sin  A 

=  9.76800 

colog  sin  Q 

=  0.11172 

log  a 

=  0.11828 

a 

=  1.313. 

5.   Given  find 

a  =  0.917,  ^  =  132°  18^  27^^ 

6  =  0.312,  B=    14°34^24^^ 

C  =  33°  7^  9^^  c  =  0.67748. 

a  +  b=  1.229 
a-b  =  0.605 
^  +  5  =  146°52^5K^ 
^(^  +  ^)=    73°  26^  25^^ 
i(A-B)=    58°  52^   1^^ 
^  =  132°  18^  27^^. 
5=    14°  34^  24^^. 

log(a-&)         =    9.78176-10 
logtanK^+^)  =  10.52674 
colog  {a+b)  =    9.91045-10 

log  tan  J(^-5)=  10.21895 

^(^-.S)=58°52^  V^. 


logS 
log  sin  C 
colog  sin  B 

logc 
c 


9.49415  - 10 

9.73750 

0.59925 

9.83090  - 10 
0.67748. 


A  =  118°  55^  49^^. 
(7=   45°4F35^^ 

log  {a-c)  =  0.39811 

log  tan  J  (il  +  C)  =  0.86968 

colog  (a  +  c)  =  8.60330 


10 


6.   Given  find 

a  =  13.715,  ^  =  118°55^49^^ 

c=  11.214,  C=  45°4V35^^ 

B  =  15°  22^  36'^ ;  b  =  4.1554. 

a -c=  2.501. 
a  +  c  =  24.929. 

A  +  C=  164°  37^  24^^ 
i{A+C)=   82°  18^  42^^ 


log  tan  H^  -  C)  =  9.87109 

i{A-C)  =  36°  37^  7''. 


log  sin  B 
log  a 
colog  sin  A 

log  6 
& 

7.   Given 
b  =  3000.9, 
c  =  1587.2, 
A  =  86°  4^  4^^ ; 


K^-C)=    36°  37^    7 


// 


=  9.42352 
=  1.13720 
=  0.05789 

=  0.61861 
=  4.1554. 

find 
5  =  65°  13^  5F^ 

(7=28°  42^    5^^ 
a  =  3297.2, 


6 +  c  =  4588.1 
6 -c=  1413.7 
B  +  C=dS°  55^  56^^ 
|(.g  +  (7)  =  46°  57^  58^^ 

J(5-C)=18°15^53^^ 
C=  28°  42^  5^^. 
5  =  65°  13^  51^^ 

log(6-c)  =3.15036 

colog  {b  +  c)  =  6.33837  -  10 

log  tan  I  (5 +(7)  =  0.02983 

log  tan  1(5  -  (7)  =  9.51856 

i(5-(7)  =  18°15^53/^. 


log& 
log  sin  A 
colog  sin  B 

log  a 
a 


3.47726 
9.99898 

:  0.04191 
3.51815 

■■  3297.2. 


TEACHERS     EDITION. 


95 


8.   Given 

find 

a  =  4527, 

^  =  68°  29^  15^^ 

b  =  3465, 

B  =  45°  24^  18^^, 

C=  66°  6^  27^^ 

;     c  =  4449. 

a  +  b=  7992. 
a  -  6  =  1062. 
^  +  5=  113°  53^33^^ 
1{A  +  B)=    56°  56^47^^ 

J(^- 

-B)=    11°  32^28^^ 
^=    68°29a5^^ 

5=    45°24a8^^ 

log  (a  -  b) 
colog  (a  +  b) 
log  tan  J  (^  - 

=    3.02612 
=   6.09734-10 
\-B)  =  10.18659 

log  tan  ^{A- 

-B)=    9.31005 
-B)  =  11°  32^  28^^. 

log  sin  Q 
colog  sin  A 
log  a 

=  9.96109 
=  0.03136       • 
=  3.65581 

logc 
c 

=  3.64826 
=  4449. 

9,   Given 
a  =  55.14, 
6  =  33.09, 
C=  30°  24^  ; 

find 
A  =  117°  24^  33^^ 
B=    32°  IF  27^/ 
c  =  31.431. 

a  +  5  =  88.23 
a  -  6  =  22.05 
A-\-B  =  149°  36^ 
^(^  +  ^)=    74°  48^ 

i(^ 

-B)=    42°36^33/^ 
^  =  117°24^33^^ 

B==   32°1F27^^ 

log  (a -6)  =    1.34341 

colog  (a  +  &)  =    8.05438  -10 

log  tan  1{A  +  B)  =  10.56592 


log  tan  ^{A- 

-B) 
-B) 

=    9.96371 
=  42°  36^  2>V^. 

log& 
log  sin  0 
colog  sin  B 

=  1.51970 
=  9.70418 
=  0.27348 

logc 
c 

=  1.49736 
=  31.431. 

10.   Given  find 

a  =  47.99,  ^  =  2°  46^    8^', 

b  =  33.14,  .5=1°  54^  42^^, 

C=  175°  19^  10^^  ;    c  =  81.066. 

a +  5  =  81.13 
a-b  =  14.85 
^  +  5  =  4°  40^  50^^ 
|(^  +  ^)  =  2°  20^  25^^ 

I  {A-B)  =  0°  25^  43^^ 
A  =  2°W  8^\ 
^  =  1°  54^  42^^ 

log  (a -J)  =1.17173 

colog  (a  +  b)  =  8.09082  -  10 

log  tan  J  [a  +  B)  =  8.61138 
log  tan  H^  -  ^)  =  7.87393 

m-B)  =  0°  25'  43^^ 


log  6 

=  1.52035 

log  sin  0 

=  8.91169 

colog  sin  B 

=  1.47680 

logc 

=  1.90884 

c 

=  81.066. 

11.  If  two  sides  of  a  triangle  are 
each  equal  to  6,  and  the  included 
angle  is  60°,  find  the  third  side. 


96 


TRIGONOMETRY. 


Since 


e 

a 

=  h, 

A 

=  B, 

A 

+  B 

=  120°. 

■.A 

=  5  =  60° 

log 

a 

=  0.77815 

log 

sin 

C 

=  9.93753 

colog 

sin 

B 

=  0.06247 

logc 

=  0.77815 

c 

=  6. 

12.   If  two  sides  of  a  triangle  are 
each  equal  to  6,  and  the  included 
angle  is  120°,  find  the  third  side. 
A  +  B  =  60°, 
.\A  =  B  =  30°, 
a  =  6  =  5. 

log  a        =0.77815 

log  sin  (7=9.93753 

colog  sin  A  =  0.30103 

log  c         =  1.01671 
c  =  10.392. 


13.  Apply  Solution  I.  to  the  case 
in  which  a  =  b  or  the  triangle  is 
isosceles. 

If  a  =  6  and  the  A  is  isosceles, 
the  angles  A  and  B  are  equal,  be- 
ing opposite  the  equal  sides. 

Now  as  a  =  6  and  A=  B,  the 
formula 

tan  J  (A-B)  =  ^^  X  tan  J  (A+B) 
a+b 

will  become 

tan  J(0)  =  Oxtan  J(2^). 
0  =  0. 

14.  If  two  sides  of  a  triangle  are 
10  and  11,  and  the  included  angle  is 
50°,  find  the  third  side. 


a  +  b  ^21 
a-b  =  l 
A  +  B  =130° 
UA  +  B)=    65° 


^{A-B)=  5°  49^  51^^ 
A=  70°49^5F^ 
B  =    59°  10^    9^\ 


log  (a -5)  =    0.00000 

colog  (a +  6)  =    8.67778-10 

log  tan  1(^  +  5)  =  10.33133 
log  tan  J  (J. -^)=    9.00911 

^{A-B)  =  5°  4:9'  5V^. 


logb 
log  sin  0 
colog  sin  B 

logc 
c 


1.00000 

9.88425 
0.06617 

0.95042 
8.9212. 


15.  If  two  sides  of  a  triangle  are 
43.301  and  25,  and  the  included 
angle  is  30°,  find  the  third  side. 

a +  5  =  68.301. 
a-5  =  18.301. 
^  +  ^  =  150° 
UA  +  B)=    75° 

^{A-B)=   45° 
A  =  120°. 
B=   30°. 
.'.  in  isos.  triangle  ABQ 
c  =  6  =  25. 

log(a-&)  =    1.26247 

colog  (a +  6)  =    8.16557-10 

log  tan^(^  +  ^)  =  10.57195 

log  tan  J  (^-^)=^    9.99999 
^{A-B)  =  45°. 


TEACHERS     EDITION. 


97 


16.  In  order  to  find  the  distance 
between  two  objects  A  and  B  sep- 
arated by  a  swamp,  a  station  C  was 
chosen,  and  the  distances  CA  =  3825 
yards,  CB  =  3475.6  yards,  together 
with  the  angle  ACB  =^  62°  SV,  were 
measured.  Find  the  distance  from 
^  to  B. 

b  +  a  =  7300.6 

b-a  =  Sid  A 
B+  A  =  117°  29^ 
i{B  +  A)=    58°  44^  30^^ 


B  = 

A  = 


4°  30^  30^^ 
63°  15^ 
54°  14^ 


log(&-a)  =    2.54332 

colog(6  +  a)  =    6.13664-10 

log  tan  ^(5  +  ^)  =  10.21680 
log  tan  J  (5  -  ^)  =    8.89676 

H^  -  ^)  =  4°  30^  30^^ 


log  6 
log  sin  C 
colog  sin  B 

logc 
c 


3.58263 
9.94799 
0.04916 
3.57978 
3800. 


17.  Two  inaccessible  objects  A 
and  B  are  each  viewed  from  two 
stations  C  and  D  562  yards  apart. 
The  angle  ACB  is  62°  12^  BCD 
41°  8^  ABB  60°  49^  and  ADC 
34°  5V ;    required  the  distance  AB. 

e 


In  triangle  ACD 

A  =  1S0°-{C+D) 

=  41°  49^ 

b    ^  sin  34°  51^ 
562     sin  41°  49^' 

.    ^  _  562  sin  34°  5V 
sin  41°  49^ 

log  562  =  2.74974 

log  sin  34°  51^  =  9.75696 
colog  sin  41°  49^  =  0.17604 
log  b  =  2.68274 

b  =  481.65. 

In  triangle  CBD 

B  =  1S0°-{C+D) 

=  43°  12^. 

a    ^  sin  95°  40^ 
562     sin  43°  12^' 

_  562  cos  5°  40^ 

sin  43°  12' 

log  562  =  2.74974 

log  cos  5°  40'    =  9.99787 

colog  sin  43°  12^  =  0.16460 


log  a 


=  2.91221 
=  816.98. 


In  triangle  ACB 

tan  }iA-B)=  "*— ^  x  tan  | (A+B) 
a  +  b 

H^  +  ^)  =  Hi80°-C) 

=  58°  54'. 
a- 6  =  816.98 -481.65 

=  335.33. 
a +  6  =  816.98 +481.65 

=  1298.63. 


98 


TRIGONOMETRY. 


log  (a—  b)  =    2.52547 

colog  {a  +  b)         =    6.88651  -10 
log  tan  i(^  +  ^)=  10.21951 


logtani(.4-5)=    9.63149 

i(^-^)=  23°  10^26'^. 
^  =  82°    4^26^A 


log  a 

log  sin  C 
colog  sin  A 

logc 
c 


2.91221 
9.94674 
0.00418 

2.86313 
729.68. 


18.  Two  trains  start  at  the  same 
time  from  the  same  station,  and 
move  along  straight  tracks  that 
form  an  angle  of  30°,  one  train  at 
the  rate  of  30  miles  an  hoar,  the 
other  at  the  rate  of  40  miles  an 
hour.  How  far  apart  are  the  trains 
at  the  end  of  half  an  hour  ? 


< 

A 
i{A 

I  +  6  =  35 
1-6  =  5 
+  B=  150° 
+  5)=    75° 

H^ 

-B)=    28°    ¥ 
B=    46°  56^ 
^  =  103°    4^ 

log  (a  -  b) 
colog  (a  +  b) 
log  tan  J  {A 

=    0.69897 
=    8.45593-10 
+  B)  =  10.57197 

log  tan  i  {A 

■1  K^ 

~B)=    9.72687 
-B)  =  28°  V. 

log  6 
log  sin  C 
colog  sin  B 

=  1.17609 
=  9.69897 
=  0.13634 

logc 
c 

=  1.01140 
=  10.266. 

19.   In  a  parallelogram  given 

the 

a  -  6  =  0.5 

two  diagonals  5  and  6,  and  the 

an- 

a  +  6  =  5.5 

gle  that  they  form  49°  18^.     Find 

^  +  ^=130°  42^ 

the  sides. 

^{A+B)=    65°  21^ 

In  the  parallelogram  ABDE 

^{A-B)=    11°  12^  20^^ 

let                EB  =  6, 

A=    76°33^20^^ 

AD^b, 

^=54°   8^40^^. 

and        Z  BCA  =  49°  18^. 

log 

(«- 

-6)             =    9.69897-10 

In  triangle  ACB 

colog 

{a  +  b)             =    9.25964-10 

let                 .5C=o  =  3, 

log 

tan 

i(^  +  ^)  =  10.33829 

^C=6=2.5. 

log 

tan 

1{A-B)=    9.29690 

Find        AB  =  c. 

i  (^  -  J5)  =  11°  12^  20^^ 

teachers'  edition. 


99 


log  a 

0.47712 

log  a                      =  0.47712 

colog  sin  J.                = 

0.01207 

colog  sin  A               =  0.34238 

log  sin  Q                = 

9.87975 

log  sin  c                  =  9.87975  -  10 

logc 

c                 =AB  = 

0.36894 

log  G                       =  0.69925 

2.3385. 

c                  =EA  =  5.0032. 

20.  In  a  triangle  one  angle  equals 

In  triangle  AEQ 

139°  54^,  and  the  sides  forming  the 
angle  have  the  ratio   5  :  9.      Find 
the  other  two  angles. 

EC= 

-  a  =  S, 

AC  = 

-.  h  =  2.5, 

a  =  9 

ZACE  = 

=  130°  42^ 

6  =  5 

A  +  E  = 

=  49°  18^ 

a +  6=  14 

K^  +  ^)  = 

-  24°  39^ 

a-&  =  4 

H^-^)  = 

=    2°  23^  20^^ 

^  +  5  =  40°    6^ 

A  = 

=  27°    2^20^'. 

^{A  +  B)  =  20°    y 

1{A-B)^    5°  57^  10^^ 
^  =  26°    OMO^^ 

log  (a  -  6) 

=  9.69897-10 

^  =  14°    5^50^^ 

colog  (a  +  b) 

=  9.25964  - 10 

log  {a-h)             =  0.60206 

log  tan  ^{A  +  E)  = 

=  9.66171 

colog  (a +  6)             =8.85387-10 

log  tan  n^  -  ^)  = 

=  8.62032 

log  tan  ^{A  +  B)=  9.56224 

i(^-^)  = 

=    2°  23^  20^^ 

log  tan  H^-^)  =  9.01817 

^  = 

.  27°    2^  20^^ 

*(^-^)=5°57M0^^ 

Exercise  XVI.     Page  67. 


1.    Given  a  =  51,  b  = 

65, 

C  =  20; 

colog  s           =    8.16749  -  10 

find  the  angles. 

a=    51 
b=    65 

colog  {s-a)  =    8.76955  -  10 
log  {s-b)  =   0.47712 

c=    20 

log  {s-c)  =    1.68124 

2s=136 

2)19.09540-20 

s=    68. 
s-a=    17. 
s-b=     3. 

log  tan  ^^=    9.54770 

J^  =  19°26^24^^ 

8-c=   48. 

A  =  38°  52^  48^^ 

100 


TRIGONOMETRY. 


colog  s  =    8.16749-10 

colog  (s-b)  =    9.52288  -  10 
log  (s-  a)  =    1.23045 
log  (s-c)  =    1.68124 

2)20.60206-20 
log  tan  1 5  =  10.30103 

J  ^  =    63°  26^    6'\ 
B  =  126°  52^  12^^ 
A^-  B  =  165°  45^. 
.-.  (7=    14°  15^ 

2.   Given  a  =  78,  6  =  101,  c  =  29  ; 
find  the  angles. 

a=    78 

&  =  101 

c  =  _29 
2s  =  208 

s  =  104. 
s  —  a=  26. 
s-h=  3. 
s  —  c=    75. 

colog  s  =    7.98297-10 

colog  (s  -  a)  =    8.58503  -  10 
log(s-5)  =    0.47712 
log  \s-c)  =    1.87506 

2)18.92018-20 
logtanJ^=    9.46009 

J^  =  16°    5^27^^ 
A  =  32°  10^  54^^ 

colog  s  =    7.98297-10 

colog  (s  -  5)  =    9.52288  -  10 
log(s-a)  =    1.41497 
log  (s  -  c)  =    1.87506 

2)20.79588-20 
logtan  1^  =  10.39794 

^B=    68°  11^  55^^ 


^=]36°23^50^^ 

^  +  5  =  168°  34^  44^^. 

.-.  C=    11°  25^  16^^ 

3.   Given  a  =  111,  5  =  145,  c  =  40; 
find  the  angles. 

a  =  111 
5  =  145 
c=    40 

2  s  =  296 

s  =  148. 

s  —  a=   37. 

s  -  5  =     3. 

s-c  =  108. 

colog  s  =  7.82974  -  10 

colog  (s- a)  =  8.43180-10 

log(s-5)  =  0.47712 

log  {s-c)  =  2.03342 

2)18.77208-20 
logtanJ^=    9.38604 

^J.  =  13°40^16^^ 
A  =  27°  20^  32^^. 

colog  s  =  7.82974-10 

log(s-a)  =  1.56820 

colog  (s  -  5)  =  9.52288  -  10 

log  {s-c)  =  2.03342 

2)20.95424-20 
log  tan|5=  10.47712 

^B=    71°  33^  54^^. 

^  =  143°    7M8^^. 

5  +  ^  =  170°  28^  20^^. 

.-.  C=     9°  3F  40^^. 


4.  Given  a  =  21,  6  =  26,  c  =  31  ; 
find  the  angles. 


TEACHERS     EDITIOX 


101 


a  =  21 
6=26 
c  =  31 

2s  =  78 

s  =  39. 

s-a=18. 

s-6  =  13. 

s-c=    8. 


colog  s 


=  8.40894-10 

colog  (s-a)  =  8.74473  -  10 

log  (s-h)  =  1.11394 

log  (s  -  c)  =  0.90309 

2)19.17070-20 
logtani^=    9.58535 

.-.  A  =  42°  6^  13^^ 

colog  s  =  8.40894-10 

log  {s-a)=  1.25527 

colog  (s  -  6)  =  8.88606  -  10 

log  {s-c)  =  0.90309 

2)19.45836-20 
logtaiiJ^=    9.72668 

.-.  B  =  56°  6^  36^^ 
^  +  ^  =  98°  12M9^^ 
.-.  C=  81°  47^  11^^ 


5.   Given  a  =  19,  5  =  34, 
find  the  angles. 

a=  19 
&=  34 
c  =  _49 

2s=102 

s=    51. 

s-a'=    32. 

s-h=    17. 

8-C=      2. 


49 


colog  s  =  8.29243  -  10 

colog  (s-a)  =  8.49485  -  10 

log  (s  -  5)  =  1.23045 

log  (s-c)  =  0.30103 

2)18.31876-20 

logtanJ^=  9.15938 

M  =  8°  12^  48^^. 


A 

= 

16°  25^  36 

v^ 

colog  s 



8.29243  - 

10 

colog  (s  - 

-h) 

= 

8.76955  - 

10 

log  (s  - 

-c) 

= 

0.30103 

log  (s  - 

-a) 

= 

1.50515 

lo^ 


2)18.86816-20 
tan  ^5=  9.43408 
J5=  15°  12^. 
B  =   30°  24^ 
.-.  C=  133°  10^  24^^ 


6.   Given  a  =  43,  5  =  50,  c- 

find  the  angles. 

=  57 

a=    43 

6=    50 

c=    57 
2s  =150 

s=    75. 

s-a=    32. 

s-b=    25. 

s-c=    18. 

colog  s            =    8.12494- 

-10 

colog  {s~a)  =    8.49485  - 

-10 

log(s-6)  =    1.39794 

log  (s-c)  =    1.25527 

2)19.27300- 

-20 

logtanJ^=  9.63650 

J^  =  23°  24^47^^. 
A  =  46°  49^  35^^ 


102 


TRIGONOMETRY. 


cologs  =    8.12494-10 

log(s-a)=    1.50515 
colog  (s  -  6)  =    8.60206  -  10 
log  {s-c)=    1.25527 

2)  19.48742  -  20 
logtanJ^=    9.74371 

J5  =  28°59^52^^ 

B  =  57°  59^  44^^. 

.-.  (7=  75°  10^  41^^ 

7.   Given  a  =  31,  5  =  58,  c  =  79  ; 
find  the  angles. 

a=  37 
6=  58 
c=_79 

2s  =  174 

s=   87. 

s  —  a=    50. 

s-b=    29. 

s-e=     8. 

colog  s  =    8.06048  -  10 

colog  {s-a)  =    8.30103  -  10 
log(s-&)  =    1.46240 
log  (s-c)  =    0.90309 

2)18.72700-20 
logtan^^=    9.36350 

J^  =  13°0M4^^. 
A  =  26°  0^  29^^. 

colog  s  =  8.06048  -  10 

log(s-a)=  1.69897 

colog  (s  -  6)  =  8.53760  -  10 

log  (s  -  c)  =  0.90309 

2)19.20014-20 

logtanJ5=    9.60007 

^B=    21°42M0^^ 

B  =    43°  25^  20^^. 

.-.  C=  110°  34^  11^/. 


8.   Given  a  =73,  6  =  82,  c  =  91; 
find  the  angles. 

a=    73 

&=    82 

c=    91 

2s  =  246 

s  =  123. 

s  —  a  =    50. 

s-b=    41. 

s  —  c=    32. 

colog  s  =  7.91009  -  10 

colog  (s-a)  =  8.30103  -  10 

log(s-6)  =  1.61278 

log  (s  -  c)  =  1.50515 


2) 

19.32905  - 

-20 

log  tan 

i^ 

9.66453 

i^ 

= 

24°  47^  29^^. 

A 

= 

49°  34^  58 

// 

colog  s 

= 

7.91009  - 

-10 

log  (s  - 

-a) 

= 

1.69897 

colog  (s  - 

-b) 

= 

8.38722- 

-10 

log  (s  - 

-c) 

= 

1.50515 

2) 

19.50143  - 

■20 

log 


tan|^=  9.75072 

^B=  29°23^29^^ 

B  =  58°  46^  58^^. 

.-.  C=  71°  38^    4^^. 


9.   Given  a  =  14.493,  b  =  55.4363, 

e  =  66.9129  ;  find  the  angles. 

a=    14.493 

b=   55.4363 

c=   66.9129 

2s  =136.8422 

s=    68.4211. 

s-a=    53.9281. 

s-&=    12.9848. 

s-c=     1.5082. 


TEACHERS     EDITION. 


103 


cologs  =  8.16481-10 

colog(s-a)  =  8.26819-10 

log(s-6)  =  1.11344 

log  (s-c)  =  0.17846 

2)17.72490-20 


log  tan  J  A 


8.86245 


A  =  8°  20^ 

cologs  =  8.16481-10 

log(s-a)  =  1.73181 

colog  (.s  -b)  =  8.88656  -  10 

log  (s-c)  -  0.17846 

2)18.96164-20 
logtaiiJ5=    9.48082 
^B=    16°  50^ 
B  =    33°  40^ 
.-.(7=138°. 

10.  Given  a=y/5,b  =  V6,  c  =  V7] 
find  the  angles. 

a  =V5  =  2.2361 

6  =V6  =  2.4495 

c=V7  =  2.6458 

2s  =7.3314 

s  =  3.6657. 

s-a  =  1.4296. 

8-6  =  1.2162. 

s-c  =  1.0199. 

log  (s-b)  =  0.08500 

log  (s-c)  =  0.00856 

colog  s  =  9.43585  -  10 

colog  (s-a)  =  9.84478  -  10 

2)19.37419-20 
logtanJ^=    9.68709 

|^  =  25°56^36^^ 
A  =  51°  53^  12^^ 


colog  (s  - 

-b)  = 

9.91500- 

10 

log  (s  - 

-c)  = 

0.00856 

colog  s 

= 

9.43585  - 

10 

log  (s  - 

-a)  = 

2) 

0.15522 

19.51463  - 

20 

log  tan 

15  = 

9.75732 

i5  = 

29°  45^  54 

// 

B  = 

59°  31^  4S 

// 

.:C= 

68°  35^ 

11.   Given 

a  =  6, 

6  =  8,  c  = 

=  10 

find  the  angles. 

a  = 

6. 

6  = 

8. 

c  = 

10. 

s  = 

12. 

s 

—  a  = 

6. 

s 

-6  = 

4. 

s 

—  c  = 

2. 

colog  s  =    8.92082  -  10 

colog  (s-a)  =    9.22185-10 
log  (s  -  6)  =    0.60206 
log  (s-c)  =    0.30103 

2)  19.04576  -  20 
log  tan  I  ^=    9.52288 

i^  =  18°26^    6^^ 
A  =  36°  52^  12^^. 
Since  this  is  a  right  triangle, 
C=  90°. 
B  =  90°  -  A 


=  53 


;c  T/ 


48^^. 


12.   Given   a  =  6,   6  =  6,  c  =  10; 
find  the  angles. 

a=  6 
6=  6 
c  =  10 

2s=22 


104 


TRIGONOMETRY. 


s  =  ll. 
s  —  a=  5. 
s  -  6  =  5. 
s  —  c  =    1. 

colog  s  =    8.95861  -  10 

colog  {s-c)  =  0.00000 
log  (s  -  6)  =  0.69897 
log  (s  -  a)  =    0.69897 

2)  20.35655  -  20 
log  tan^  (7=  10.17828 

^C=    56°  26^  33^^. 
C=  112°  53^    6'\ 

Since  this  is  an  isosceles  triangle, 
^  =  ^=1(180°-  C) 
=  33°  33^  27^^. 

13.  Given   a  =  6,   &  =  6,   c  =  6  ; 
find  the  angles. 

The    triangle  is  equilateral  and 
also  equiangular. 

.-.  A  =  B=C=^of  180°  =  60°. 

14.  Given  a  =  6,   6  =  5,  c  =  12  ; 
find  the  angles. 

The  sum  of  the  two  sides  a  and  b 
is  less  than  the  side  c. 

.•.  the  triangle  is  impossible. 

15.  Given    a  =  2,    h  =V6,    c  = 
VS  —  1  ;  find  the  angles. 

a  =  2 

6  =  V6  =  2.4495 

c=  VS- 1  =  0.732 

2  6  =  5.1815 

5  =  2.5908. 

s-a  =  0.5908. 

s- 6  =  0.1413. 

s-c=  1.8588. 


log(s-a)  =    9.77144-10 
log(5-&)  =    9.15014-10 
log  {s-c)  =    0.26923 
colog  s  =    9.58656  -  10 

log  r2  =  18.77737  -  20 

logr  =    9.38869-10 

log  tan  ^  J.  =  9.61725 
logtan|5  =  10.23855 
logtan^(7=    9.11946 


iA=    22°  30^ 

^B=    60°. 

i  C  =      7°  30^. 

J.=    45°. 

B  =  120°. 

C=    15°. 

16.   Given    a  =  2,   6  =  Ve, 

c 

\/3  +  1  ;  find  the  angles. 

a  =  2 

6  =  V6  =  2.4495 

c  =  V3  +  1  =  2.732 

2s=  7.1815 

s  =  3.5908 

s-a  =  1.5908 

s-&  =  1.1412 

s-c  =  0.8588 

log  {s-a)  =   0.20162 

log  (s  -  5)  =   0.05740 

log  (s-c)  =    9.93385  - 

10 

colog  s            =   9.44481  - 

10 

logr2           =19.63768- 

20 

logr           =   9.81884- 

10 

logtanJJ.  =  9.61721. 
log  tan  J5  =  9.76146. 
logtan  J  (7=9.88494. 


TEACHEES     EDITION. 


105 


hA  = 

22°  30^. 

J5  = 

30°. 

iC  = 

37°  30^ 

A  = 

45° 

B  = 

60° 

C  = 

75° 

17.  The  distances  between  three 
cities  A,  B,  and  C  are  as  follows : 
AB  =  165  miles,  AC=  72  miles,  and 
BC  =  185  miles.  B  is  due  east  from 
A.  In  what  direction  is  Cfrom  A  ? 
What  two  answers  are  admissible  ? 

a  =185 
6=  72 
c  =  165 

2s  =  422 

s  =  211. 

s-a=    26. 

s-6  =  139. 

s-c=    46. 

cologs  =    7.67572-10 

colog  (s-a)  =    8.58503  -  10 
log-(s-6)  =    2.14301 
log  [s-c)  =    1.66276 

2)  20.06652  -  20 
logtan^^  =  10.03326 

iA  =  i1°  IF  30^^. 
A  =  94°  23^ 

Angle  BAC=  94°  23^  Subtract 
90°  of  the  quadrant  U  to  N,  and 
we  obtain  4°  23^  W.  of  N. 

But  C  may  be  to  the  southward 
of  A.  Hence  two  answers  are  ad- 
missible: W.  of  N.  or  W.  of  S. 

18.  Under  what  visual  angle  is 
an  object  7  feet  long  seen  by  an 


observer  whose  eye  is  5  feet  from 
one  end  of  the  object  and  8  feet 
from  the  other  end  ? 

a=    5 

&=    8 

c  =  _7 

2s  =  20 

s  =  10. 

s  —  a=    5. 

s-h=    2. 

s  —  c=    3. 

cologs  =    9.00000-10 

colog  (s  -  a)  =    9.30103  -  10 
log(s-&)  =    0.30103 
log(s-e)   =    0.47712 

2)19.07918-20 
logtan|^=    9.53959 

J^  =  19°    6^24^^. 
A  =  38°  12^  48^^. 


colog  s 

= 

9.00000  - 

-10 

log  (s  - 

-a)  = 

0.69897 

colog  (s  - 

-6)  = 

9.69897- 

-10 

log  (s  - 

-c)  = 

0.47712 

2)_ 

19.87506  - 

-20 

log  tan 

hB  = 

9.93753 

iB  = 

40°  53^  36^^ 

B  = 

81°  47^  12^^. 

.\C= 

60°. 

19.  When  Formula  [28]  is  used 
for  finding  the  value  of  an  angle, 
why  does  the  ambiguity  that  occurs 
in  Case  II.  not  exist? 

When  Formula  [28]  is  used  for 
finding  the  value  of  an  angle,  the 
ambiguity  that  occurs  in  Case  II. 
does  not  exist  because  the  sides 
are  all  known  and  the  angle  can 


106 


TEIGONOMETRY. 


have  but  one  value ;  while  in  Case 
II.  the  side  opposite  the  angle  is 
not  known,  and  may  have  two  val- 
ues ;  therefore  the  angle  also  may 
have  two  values. 

20,  If  the  sides  of  a  triangle  are 
3,  4,  and  6,  find  the  sine  of  the 
largest  angle. 

a  =    3 

h=    4 
c  =  _6 

2s  =  13 
5  =  6.5. 


s 

—  a 

=  3.5. 

s 

-h 

=  2.5. 

s{s- 

-c) 

=  3.25. 

log 

{s- 

-a) 

=    0.54407 

log 

(s- 

-h) 

=    0.39794 

colog 

S{S' 

-c) 

=    9.48812- 
2)20.43013- 

10 
20 

log 

tan 

\G 

=  J  0.21507 

\G 

=    58°38^2£ 

f^. 

C=117°16^50^^ 
log  sin  C     =  9.94879. 
sin  C  "  =  0.88877. 


21.  Of  three  towns  A,  B,  and  C, 
A  is  200  miles  from  B  and  184  miles 
from  (7,  B  is  150  miles  due  north 
from  C;  how  far  is  A  north  of  (7  ? 

a  =  150  s  =  267. 


6  =  184 

s-a  =  117. 

c  =  200 

s-h=    83. 

2s  =  534 

s-c=    67. 

colog  s  =    7.57349  -  10 

colog  (s-c)  =    8.17393-10 
log  (s  -  a)  =    2.06819 
log  (s-b)  =    1.91908 

2)19.73469-20 
Iogtan^^=    9.86735 


itl 


a' 

G 

J^  =  36°22^58^^ 

A  -  72°  45^  56^^ 

Draw  ±  from  A  to  BC.     To  find 
a^  (part  cut  off  by  ±  on  BC  from  c). 

a^=  b  cos  C. 

log  b  =  2.26482 

log  cos  C  =9.47171 

log  a^  =  1.73653 

a'  =  54.516. 


TEACHERS     EDITION. 


107 


Exercise  XVII.     Page  69. 


1.  Given  a  =  4474.5,  b  =  2164.5, 
C=  116°  30'  20^' ;  find  the  area. 

F=  Ja&sin  C. 

log  a  =  3.65075 

log  b  =  3.33536 

colog  2  =  9.69897  -  10 

log  sin  C    =  9.95177 

log  F         =  6.63685 
F  =4333600. 

2.  Given     b  =  21.66,    c  =  36.94, 
A  =  66°  4'  19^'  ;  find  the  area. 

F=^bcsmA. 


log  6 

=  1.33566 

logc 

=  1.56750 

log  sin  A 

=  9.96097 

\og2F 

=  2.86413 

2F 

=  731.36. 

F 

=  365.68. 

3.  Given  a  =  510,  c  =  173,  B  = 
162°  30'  28''  ;  find  the  area. 

log  a  =  2.70757 

log  c  =  2.23805 

log  sin  B  =  9.47795 

colog  2  =  9.69897  -  10 

log  F  =  4,12254 

F  =13260. 

4.  Given  a  =  408,  6  =  41,  c  =  401; 
find  the  area. 

a  =  408 
6=  41 
c=401 

2s  =  850 
s  =  425. 


s  —  a  =  17. 
s-b  =  384. 
s-c=    24. 

log  s  =  2.62839 

log(s-a)  =1.23045 
log  {s-b)  =  2.58433 
log(s-c)  =1.38021 


2)7.82338 

log  F          =  3.91169 

F               =  8160. 

5.    Given  a  =  40,  6  =  13,  c  = 
id  the  area. 

a  =  40 

=  37; 

6  =  13 

c  =  37 
2s  =  90 

s  =  45. 

s  —  a  =    5. 

s-b  =  32. 

s-c=    8. 

log  s  =  1.65321 

log  (s  -  a)  =  0.69897 
log  (s  -  6)  =  1.50515 
log  (s-c)  =  0.90309 

2)4.76042 
log  F  =  2.38021 
F  =  240. 

6.  Given  a  =  624,  5  =  205,  c  =  445; 
find  the  area. 

a=  624 
6=  205 
c=    445 


2s  =  1274 


108 


TRIGONOMETRY. 


s=  637. 

s  —  a=  13. 

s-b=  432. 

s-c=  192. 

log  s  =  2.80414 

log  {s-a)  =  1.11394 
log  Is-b)  =  2.63548 
log  {s-c)  =  2.28330 

2  log  i^  =  8.83686 
log  F  =  4.41843. 
F  =  26208. 


7.  Given  h  =  149,  ^  =  70°  42^  30^^ 

-  39°  18^  28^^ ; 

find  the  area. 

A 

=  70°  42^  30^'' 

B 

=  39°  18^  28^^ 

.'.C 

=  69°  59^    2^^ 

log  6 

=  2.17319 

colog  sin  B 

=  0.19827 

log  sin  A 

=  9.97490 

log  a 

=  2.34636 

log  a 

=  2.34636 

colog  sin  yl 

=  0.02510 

log  sin  C 

=  9.97294 

.     logc 

=  2.34440 

colog  2 

=  9.69897  - 10 

log  a 

=  2.34636 

log& 

=  2.17319 

log  sin  C 

=  9.97294 

logF 

=  4.19146 

F 

=  15540. 

log  c  =  2.48813 

log  sin  A   =  9.62852 
colog  a  =  7.66575  —  10 

log  sin  C   =9.78240 

C=   37°  17^  38^^. 

-•.  5  =  117°  32^  51^^ 

Or,         C  =  142°  42^  22^^ 

.'.B^=    12°    8^    7^^. 


colog  2  =  9.69897 

log  a  =  2.33425 

log  c  =■  2.48813 

loR  sin  B  =  9.94774 


10 


8.  Given  a  =  215.9,  c- 307.7,^ 
25°  9^  31^^  ;   find  the  area, 
n  <  c  and  >  c  sin  A. 
A  <  90°.     .'.  two  solutions. 


logi^ 

=  4.46909 

F 

=  29450. 

colog  2 

=  9.69897  - 

-10 

log  a 

=  2.33425 

logc 

=  2.48813 

log  sin 

B' 

=  9.32269 

logjP'' 

=  3.84404 

F' 

=  6983. 

9.  Given  6  =  8,   c  =  5,  ^  =  60°; 
find  the  area. 

F=\'bc  sin  J. 

=  1(8x5)  (0.86602) 
=  20  X  0.86602 
=  17.3204. 

10.  Given  a  =7,  c  =  3,  ^  =  60°; 
find  the  area. 

colog  a  =  9.15490  -  10 

log  c  =  0.47712 

log  sin  A  =  9.93753 

log  sin  C  =  9.56955 

C  ^  21°  47^  12^^. 

.-.  B  =  98°  12^  48^^ 


TEACHEES     EDITION. 


109 


log 

a 

=  0.84510 

log 

sin  B 

=  9.99552 

log 

sin  A 

=  0.06247 

log 

b 

=  0.90309 

b 

=  8. 

F= 

■■  J  be  sin  A 

=  J  X  24  X  W3 

=  6V3, 

or  10.3923 

11.  Given  a  =  60,  ^  =  40°  35^  12^^ 
area  =  12  ;  find  the  radius  of  the 
inscribed  circle. 


^  ac  sin  B  =  12. 


24 


log  24 
colog  a 
colog  sin  B 

log  c 

c 


a  sin  B 
=  1.38021 

=  8.22185 
=  0.18665 


10 


=  9.78875-10 
=  0.61483. 


tanH^-C) 


a 


a  +  c 


Xtan^(J.  +  C) 


=  ^^-^^^^"^  X  tan  (69°  42^  24^0- 
60.61483  ^  ^ 

log  {a-c)=  1.77368 
colog  (a  +  c)  =  8.21742  -  10 
logtani(^+g)=  0.43206 
log  tan  H^-C')  =  0-42316 


i{A-C)=  69°  IV  19^^ 
i{A+C)=  69°  42^  24^'' 
A  =139°    1M3^^ 


B 


a  sinB 


a     sin  A 

sm-d 

log  a 

=  1.77815 

log  sin 

B    =9.81331 

colog  sin 

A    =0.18331 

log  5 

=  1.77477 

b 

=  59.534. 

a=   60 

6=   59.534 

c=     0.61483 

2s  =120.14883 

s=    60.07442. 

s 

-  a  =     0.07442. 

s 

-b=     0.54042. 

s 

-c=   59.45959. 

log  {&-a)=  8.87169  -  10 

log  (s  -  5)  =  9.73274  -  10 

log  {&-c)=  1.77422 

colog  c  =  8.22131-10 

2)18.59996-20 

logr  =  9.29998-10 

r  =  0.19952. 


12.  Obtain  a  formula  for  the  area 
of  a  parallelogram  in  terms  of  two 
adjacent  sides  and  the  included 
angle. 

By  Geometry,  area  of  parallelo- 
gram =  base  X  height. 

In  this  case  =  bli. 

But  A  =  a  sin  J.. 

.•.  area  oi  LJ  ^  ab  sin  A. 


110 


TRIGONOMETRY. 


a 


13.  Obtain  a  formula  for  the  area 
of  an  isosceles  trapezoid  in  terms  of 
the  two  parallel  sides  and  an  acute 
angle. 

Let  AB  =  a. 

F=^{a  +  h)  c. 

-  =  tan  A 
P 
c  =p  tan  A. 

.-.  F=  l{a^h)x^{a—h)  tan  A 
=  \{a?-  ¥)  tan  A. 

14.  Two  sides  and  included  angle 
of  a  triangle  are  2416,  1712,  and 
30° ;  and  two  sides  and  included 
angle  of  another  triangle  are  1948, 
2848,  and  150°;  find  the  sum  of 
their  areas. 

Let  a  =  2416,  c  =  1712,  B  =  30°. 
F=  h  ac  sin  B. 


log  a           =3.38310 

log  c           =  3.23350 

colog  2           =  9.69897  - 

-10 

log  sin  B    =  9.69897 

log  F         =  6.01454 

F               =1034000.. 

Let  a^=  1948,  c'=  2848,  5^=150= 
jr/=iaVsin5^. 


P 

log  a^         =  3.28959 
log  c^  =  3.45454 

colog  2  =  9.69897  -  10 

log  sin  B^  =  9.69897 

log  F^        =  6.14207 
F^  =  1387000. 

F+F^       =2421000. 

15.  The  base  of  an  isosceles  tri- 
angle is  20,  and  its  area  is  lOO-r-VS ; 
find  its  angles. 

a  =  b. 


c  = 

=  20. 

F= 

=  100^  Vs. 

^ch  = 

100 
V3 

10  h  = 

100 
V3 

h^ 

10 
V3 

h 

=  tan  A. 

log  h 

= 

-  0.76144 

log  Jc 

= 

=  9.00000  - 

10 

log  tan  A  =  9.76144 

A  =  30°. 

B  =  30°. 

C  =  120°. 


TEACHERS     EDITION. 


Ill 


Exercise  XVIII.     Page  70. 


1.  From  a  ship  sailing  down  the 
English  Channel  the  Eddy  stone  was 
observed  to  bear  N.  33°  45^  W. ; 
and  after  the  ship  had  sailed  18 
miles  S.  67°  30^  W.  it  bore  N.  11° 
15^  E.  Find  its  distance  from  each 
position  of  the  ship. 


a  =  18  miles. 
ACE^  33°  45^ 
DCB=  67°  30°. 
ABF=  11°  15^ 
ACB  =  1S0°~{ACE  +  DCB) 

=  78°  45^ 
CBD  =  90°  -DCB 

=  22°  30^ 
ABC=  90°  -  {CBD  +  ABF) 

=  56°  15^ 
.^^C=45°. 

6_sin_S         c  _  sin  C 
a     sin  J. 


sin  A 


log  a 

=  1.25527 

log  sin  B 

=  9.91985 

colog  sin  A 

=  0.15051 

logb 

=  1.32563 

b 

=  21.166. 

log  a 

=  1.25527 

log  sin  C 

=  9.99157 

colog  sin  A 

=  0.15051 

logc 

=  1.39735 

c 

=  24.966. 

2.  Two  objects,  A  and  B,  were 
observed  from  a  ship  to  be  at  the 
same  instant  in  a  line  bearing  N. 
15°  E.  The  ship  then  sailed  north- 
west 5  miles,  when  it  was  found 
that  A  bore  due  east  and  B  bore 
north-east.  Find  the  distance  from 
Ato  B. 

N 


W- 


E 


„'^  45°  105° /. 

^\45°  75°; 


\ 

d^ 

V 

/ 

/ 

^n60°/ 

N         / 

NS 

112 


TRIGONOMETRY. 


S'A     sin  ASS' 

log  140       =  2.14613 

SS'      sin  S'AS 

colog  sm  A    =  0.00665 

log  SS'           =  0.69897 

log  AC      =  2.15278 

colog  sin  SAS'  =  0.01506 

HCA  =  10°, 

log  sin  ASS'  =  9.93753 

MCB  =  40°, 

log  S'A         =  0.65156 

.-.  ^C5  =  40°; 

CAB  =  10°, 

AB      sin  BS'A 

.-.  ^5(7=130°. 

S'A     sm  S'BA 

4P_  -ACsin  C 

log  /S^^          =  0.65156 

sin  -B 

colog  sm  S'BA  =  0.30103 

log.4C      =2.15278 

log  sin  BS'A  =  9.84949 

log  sin  C    =9.80807 

log  AB          =  0.80208 

colog  sin  B    =  0.11575 

AB                =  6.3399. 

log  AB      =  2.07660 

• 

AB             --=  119.29. 

3.  A  castle  and  a  monument 
stand  on  the  same  horizontal  plane. 
The  angles  of  depression  of  the  top 
and  the  bottom  of  the  monument 
viewed  from  the  top  of  the  castle 
are  40°  and  80°  ;  the  height  of  the 
castle  is  140  feet.  Find  the  height 
of  the  monument. 


110=  height  of  castle. 

AB  =  height  of  monument. 
MCB  =  ^0°.  BAC  =S0°. 

HCA  =  10°.  HC    =140  ft. 

140 


AC= 


sin  A 


4.  If  the  sun's  altitude  is  60°, 
what  angle  must  a  stick  make  with 
the  horizon  in  order  that  its  shadow 
in  a  horizontal  plane  may  be  the 
longest  possible  ? 

The  shadow  of  the  stick  will  be 
the  longest  when  the  stick  is  per- 
pendicular to  the  rays  of  the  sun. 

Let  BC  represent  the  stick,  and 
^Cthe  horizontal  plane. 

B  =  90°. 

A  =  60°. 

.-.  C  =  30°. 

5.  If  the  sun's  altitude  is  30°, 
find  the  length  of  the  longest  shadow 
cast  on  a  horizontal  plane  by  a 
stick  10  feet  in  length. 

Let  a  be  a  stick  A.  to  rays  of  sun, 
and  c  be  the  longest  shadow. 

-  =  sm  A. 
c 


TEACHERS     EDITION. 


113 


log  a 
colog  sin  A 
logc 
c 


1.00000 
0.30103 
1.30103 
20. 


6.  In  a  circle  with  the  radius  3 
find  the  area  of  the  part  comprised 
between  parallel  chords  whose 
lengths  are  4  and  5.  (Two  solu- 
tions.) 


log(s-6)        =    0.30103 
log  (s-c)        =    0.30103 
cologs(s-a)       =    9.30103-10 
2)19.90309-20 
logtanJ-BOC=   9.95155-10 
J50C=  41°  48^  38^^ 
BOQ=  83°  37^  16^^ 


By  Table  VI. 

B  =  Z. 
:.  area  O  =  28.274. 
301036 


BOQ- 


1296000 
75259 


X360° 


324000 
area  sector  BOC 

75259 


X360^ 


X  28.274. 


log  75259 
log  28.274 
colog  324000 

log  area 

Area  sector  50C=  6.5675. 


324000 

=  4.87656 
=  1.45139 
=  4.48945  - 10 

=  0.81740 


In  triangle  BOO 


F=Vs{s- 

-  <z)  (s  —  6)  (s  —  c). 

log  s{s  —  a) 

=  0.69897 

log  (s  -  b) 

=  0.30103 

log  (s  -  c) 

=  0.30103 

2)1.30103 

log  if 

=  0.65052 

F 

=  4.4722. 

Area  segment  ByC=  2.0953. 

In  triangle  BOA 

tajii  DOA  = 

^      s  (s  —  o) 

0=    5 

a=   3 

d=   3 

2s  =  ll 

s=   5.5. 

s- 

-d=    2.5. 

s- 

-a=    2.5. 

s(s- 

-o)=   2.75. 

114 


TRIGONOMETRY. 


log  {s-d)       =0.39794 

log  (s  -  a)        =  0.39794 

colog  s{s-o)      =  9.56067  -  10 

2)0.35655 

logtanJDOJ.  =  0.17828 

^DOA=    56°  26^35.5^^ 
DOA  =  112°  53^  IV^, 


In  triangle  DOA 

F=  \^s{s-o){s-a){s-d). 

log  s{s-o)      =  0.43933 
log  (s  -  d) 
log  (s  -  a) 


\ogF 
F 


=  0.39794 
=  0.39794 

2)1.23521 
=  0.61761 
=  4.1458. 


DOA 


sector  DOA  = 


40^391^ 
1296000 
406391 


X  360°. 


X  28.274. 


1296000 
log  406391    =5.60894 
log  28.274     =1.45139 
colog  1296000  =  3.88739-10 

log  area         =  0.94772 
Area  sector  =  8.8658. 

Area  segment  DxA 
=  4.72. 
Area  segment  DACB 

=  areaO-[%C+i)a;^] 

=  21.4587. 
Area  segment  DAC^B^ 

=  DxA  -  B'xC^ 

=  2.6247. 


7.  A  and  B,  two  inaccessible  ob- 
jects in  the  same  horizontal  plane, 
are  observed  from  a  balloon  at  C 
and  from  a  point  D  directly  under 
the  balloon,  and  in  the  same  hori- 
zontal plane  with  A  and  B.  If  CD 
=  2000  yards,  Z  ^  CD  =  10°  15^  10^^ 
Z  BCD  =  6°  7^  20''^  Z  ADB  =  49° 
34^  50^^  find  AB. 

AD  =  DCxt&nACD, 
logtanAOD    =9,25739 
log  DO  =  3.30103 


log^D 


2.55842 


AD  =  361,76. 


DB  =  DC xts^n  BCD. 

log  DC 

=  3.30103 

log  t&n  BCD  =9.03045 

log  DB 

=  2.33148 

DB 

=  214.53. 

tan  1(5- 

-A) 

6- 

-^Xta,n^{B  +  A) 

b  +  a 


TEACHEES     EDITION. 


115 


U^  +  ^)      =  65°  12^  35^^. 

^ 

=  94°    9^33^^. 

log  (6  -  a)            =  2.16800 

log^D 

=  2.55842 

colog  (6  +  a)           =  7.23936  -  10 

colog  sin  B 

=  0.00115 

log  tan  J  (5+^)  =0.33549 

log  sm  C 

=  9.88156 

log  tan  §(^-^)  =9.74285 

log  c 

=  2.44113 

J(^-^)=28°56^58^^ 

c  =  ^5 

=  276.14. 

8.  -4  and  5  are  two  objects  whose 
distance,  on  account  of  intervening 
obstacles,  cannot  be  directly  meas- 
ured. At  the  summit  C  of  a  hill, 
whose  height  above  the  common 
horizontal  plane  of  the  objects  is 
known  to  be  517.3  yards,  Z.ACB 
is  found  to  be  15°  13^  15^^  The 
angles  of  elevation  of  C  viewed  from 
A  and  B  are  21°  9^  18^^  and  23° 
15''  34''^  respectively.  Find  the  dis- 
tance from  A  to  B. 

In  triangle  BCA,  being  a  rt.  A, 


^=sinA 

0 

sm  A 

\ogd 

=  2.71374 

colog  sin  A 

=  0.4426:2 

log  k 

=  3.15636 

h 

=  1433.4. 

In  right  triangle  (HBB 


-  =  sinB. 


a  = 


d 


sin^ 


log  d 

=  2.71374 

colog  sin  B 

=  0.40352 

log  a 

=  3.11726 

a 

=  1310. 

tan  H^  -  ^) 

-  f  -  "*  X  tan  2^  (5  +  A). 
0  +  a 

HB  +  ^) 

=    82°  23^  22.5^^ 

log  {b  -  a) 

=    2.09132 

colog  (b  +  a) 

=   6.56171-10 

logta,n^{B+A) 

=  10.87415 

log  tan  f  (5-^) 

=    9.52718 

^{B-A) 

=    18°  36^  21^^. 

B 

=  100°  59^  43.5^^. 

A 

=    63°  47^    1.5^^ 

c 

a  sin  0 

sin  A 

log  a 

=  3.11726 

log  sin  C 

=  9.41920 

.eolog  sin  A 

=  0.04714 

log  c 

=  2.58360 

€ 

=  383.35. 

SPHEEICAL  TEIGOIsrOMETET. 


Exercise  XIX.    Page  104. 


1.  The  angles  of  a  triangle  are! 
70°,  80°,  and  100°;  find  the  sides' 
of  the  polar  triangle. 

Given  A  =  70°,  B  =  80°,  C=  100°; 
to  find  a\  y,  c^. 

a^=180°-    70°  =  110°. 

6/=  180°-   80°  =  100°.  : 

c/=i80°-100°=   80°. 

2.  The  sides  of  a  triangle  are  40°, 
90°,  and  125° ;  find  the  angles  of 
the  polar  triangle. 

Given  a  =  40°,  b  =  90°,  c  =  125°  ; 
required  A^,  B\  C . 

^/=180°-   40°  =  140°. 

^/=180°-    90°=    90°. 

(7=  180°  -  125°  =   55°. 

3.  Prove  that  the  polar  of  a 
quadrantal  triangle  is  a  right  tri- 
angle. 

Let  the  triangle  ABC  be  a  quad- 
rantal triangle. 

Then  I  =  90°. 

Let  A^B^C  be  the  polar  triangle. 

B'^  h  =  180°. 

But  5=   90°. 

.-.  ^^=   90°. 

.'.  triangle  A'B^C  is  a  right  tri- 
angle. 


4.  Prove  that,  if  a  triangle  have 
three  right  angles,  the  sides  of  the 
triangle  are  quadrants. 

If  angles  A,  B,  Q  respectively, 
are  right  angles,  the  side  b  the  meas- 
ure of  angle  B,  c  the  measure  of 
angle  C,  and  a  the  measure  of  angle 
A,  are  each  =  90°  ; 

.•.  sides  of  triangle  ABC  are 
quadrants. 

5.  Prove  that,  if  a  triangle  have 
two  right  angles,  the  sides  opposite 
these  angles  are  quadrants,  and  the 
third  angle  is  measured  by  the 
number  of  degrees  in  the  opposite 
side. 

In  spherical  triangle  ABC 

let  B  =  C=  rt.  angle. 

We  are  to  prove  AC  and  AB 
quadrants,  also  that  A  is  measured 
by  the  number  of  degrees  in  BC. 

Let  A^B^C^  be  the  polar  triangle 
of  ABC. 
Now        B  +  b^=  180°, 
.-.  b'=    90°  ; 
C-fc^=  180°, 
c^=  90°. 
.•.  triangle   A^B^C^  is    isosceles, 
and    5^=90°,  (7=90°. 
But         B^+  b  =  180°. 


118 


TRIGONOMETRY. 


.-.  h=    90°. 
C'+  c  =  180°, 

.-.  c=    90°. 
.*.  5  and  c  are  quadrants. 
Now  ABCi?,  bi-rectangular. 
.'.  A  is  the  pole  of  side  BC. 
.'.  J.  is  measured  by  side  BC. 

6.  How  can  the  sides  of  a  spheri- 
cal triangle  be  found  in  units  of 
length,  when  the  length  of  the 
radius  of  the  sphere  is  known. 

By  using  the  formula  2  irB  =  C. 

For  instance,  if  the  sides  of  a  tri- 
angle were  40°,  90°,  125°,  the  sides 
in  terms  of  R  would  be 


360° 

X2  7ri2, 

90° 
360° 

X27ri2, 

125° 
360° 

X2irR. 

If  in 

this  case  i2  =  4,  the 

sides 

would  become  respectively  f  tt,  2ir, 

^TT. 

7.  Find  the  lengths  of  the  sides 
of  the  triangle  in  Example  2,  if  the 
radius  of  the  sphere  is  4  feet. 


a  =  125°. 

6=   90°. 

c  =    40°. 

circ.  =  2  ttB 

=  27rX4 

=  8  IT. 

a  =  l2f  of8,r  =  -2#T. 
&  =  /6«o  of  8T  =  27r. 
C  =  3VoOf87r  =  f7r. 


Exercise  XX.    Page  108. 


1.  Prove,  by  aid  of  Formula  [37], 
that  the  hypotenuse  of  a  right  tri- 
angle is  less  than  or  greater  than 
90°,  according  as  the  two  legs  are 
alike  or  unlike  in  kind. 

When  the  legs  are  alike  in  kind. 
By  Formula  [37] 

cos  c  =  cos  a  X  cos  h. 
If  a  and  b  are  less  than  90°,  the 
cosines  will  be  positive,   and  the 


product  a  positive  quantity ;  that 
is,  greater  than  0. 

If  a  and  b  are  greater  than  90°, 
their  cosines  will  both  be  negative, 
but  their  product  will  be  a  positive 
quantity. 

In  either  case,  cos  c  =  a  positive 
quantity. 

.•.  its  value  is  less  than  90°. 

When  the  legs  are  unlike  in  kind. 

If  a  is  greater  than  90°  and  b  less 


TEACHERS     EDITION. 


119 


than  90°,  their  cosines  have  opposite 
signs,  and  their  product  will  be 
negative. 

If  cos  c  =  a  negative  quantity, 
then  c  is  greater  than  90°. 

2.  Prove,  by  aid  of  Formula  [40], 
that  in  a  right  spherical  triangle 
each  leg  and  the  opposite  angle  are 
always  alike  in  kind. 

Formula  [40], 

cos  A  =  cos  a  X  sin  B. 

B  <  180°.     .-.  sin  B  is  positive. 

Hence  sign  of  cos  A  is  same  as 
sign  of  cos  a,  and  both  must  be 
either  greater  or  less  than  90° ;  that 
is,  alike  in  kind. 

3.  What  inferences  may  be  drawn 
respecting  the  values  of  the  other 
parts :  (i.)  if  c  =  90° ;  (ii.)  if  a  =  90° ; 
(iii.)  if  c  =  90°  and  a  =  90° ;  (iv.)  if 
a  =  90°  and  6-90°? 

(i.)  If  c  =  90°. 

0  =  cos  a  X  cos  h. 
.'.  cos  a  or  cos  6  =  0. 
Hence  a  or  6  =  90°. 

(ii.)  If  a  =  90°. 

cos  J.  =  0  X  sin  B. 
.-.  cos  ^  =  0. 
.-.  A  =  90°. 
cos  -B  =  COS  6  X  1 
=  cos  6. 
B=h. 
If      6  =  90°, 

B  =  90°  and  A  =  a. 

(iii.)  If  c  =  90°  and  a  =  90°. 
cos  A  =  COS  a  sin  B. 
A  =  a  =  90°. 


Also         C=c=  90°. 
cos  B  =  cos  6. 
B  =  h. 

(iv.)  If     a  =  90°,  6  =  90°. 
Then       A  =  90°,  B  =  90°. 

cos  c  =  0  X  0. 

.-.  c  =  90°. 

.-.  C=  90°. 

4.  Deduce  from  [37]-[42]  the  for- 
mula 

tan^  ^  6  =  tan  ^{c  —  a)  tan  ^{0  +  a). 

From  [37],  page  74,  we  have 

cos  6  =  cos  c  sec  a  ; 
whence 

1      „^o  J.     cos  a  —  cos  c 
1  —  cos  0  = -I 

COS  a 

1    .  ^^„  7>     cos  a  +  COS  c 
1  +  cos  0  = 

cos  a 
.    1  —  cos  6  _  cos  a  —  cos  c 
1  +  cos  6      cos  a  +  cos  c 

But  by  [18],  page  47, 
1  —  cos  6 


1  +  cos  6 


=  tan^  ^  6. 


And  if  in  [23]  and  [22],  page  48, 
we  write  a  and  c  in  place  of  A  and 
B  and  divide  [23]  by  [22],  we  get 

cos  a  —  cos  c 

cos  a  +  cos  c 

=  —  tan  ^{a  -j-  e)  tan  i{a  —  c) 
=  tan  J  (c  +  a)  tan  J  (c  —  a). 
.'.  tan^  J  6 

=  tan  ^  (c  +  a)  tan  f  (c  —  a). 

5.   Deduce  from  [37]-[42]  the  for- 
mula 
tan2  (45°  -^A) 

=  tan  J  (c  —  a)  cot  J  (c  +  a). 


120 


TRIGONOMETRY. 


From  [38], 
sin  A  = 


sm  a 


sm  c 

when,  operating  as  in  Example  4, 

we  have 

1  —  sin  A  _  sin  c  —  sin  a 
1  +  sin  A      sin  c  +  sin  a 

If  in  [19],  page  47,  we  substitute 
90°  -r  A  for  z,  and  remember  that 
cos  (90°  +  A)  =  —  sm  A,  [19]  reduces 
to  the  form 

^^^^  =  cot2(45°+J^) 
1  +  sm  ^ 

=  tan2(45°-J^), 

(since  45°  +  J  J.  and  45°  —  ^A  are 
complementary  angles). 

And  by  dividmg  [21]  by  [20], 
page  48,  and  writing  c  for  A  and  a 
for  B,  we  have 

sin  c  —  sin  ^4 
sin  c  +  sin  J. 

=  tan  ^{c  —  a)  cot  J  (c  +  a). 
.•.tan2(45°-J.4) 

=  tan  }{c  —  a)  cot  J  (c  +  a). 

6.  Deduce  from  [37]-[42]  the  for- 
mula 

tan^  ^  B  =  sin  (c  —  a)  esc  (c  +  a). 

From  [39],  by  operating  as  be- 

'     1  —  cos  B  _  tan  c  —  tan  a 
1  +  cos  B     tan  c  +  tan  a 
By  [18],  page  47, 

l^^^^  =  tanH5. 
1  +  cos  i? 

And  we  write  on  the  second  side 

sin  G  ■       -,  c  i.  A  sin  a    • 

■ m  place  oi  tan  c,  and  m 

cos  c  cos  a 

place   of  tan  a,   and  reducing,  we 

obtain 


tan  c  —  tan  a  _  sin  (c  —  a) 
tan  c  +  tan  a     sin  (c  +  a) 

1  p  _  sin  (c  —  g) 
sin  (c  +  a) 

sin  (c  —  a)  esc  (c  +  a). 


t&n^B 


7.   Deduce  from  [37]-[42]  the  for- 
mula 

tan^  J  c  =  —  cos  {A+B)  sec  {A—B). 

By  [42],  cos  c  =  cot  ^  cot  B 
_  cot  J. . 
tan  B' 
whence,  as  before, 

1  —  cos  c  _  tan  B  —  cot  A 
1  +  cos  c      tan  B  +  cot  A 

By  [18],  page  47, 

1  —  cos  c 


and 


tan^  I  c, 
1  +  cos  c 

tan  B  —  cot  A 
tan  B  +  cot  A   ■ 

sin  ^  cos  A 

cos  ^  sin  J. 

sm  B  cos  ^ 

cos  B  sin  J. 

_  sin  Asm  B  —  cos  J.  cos  ^ 
sin  Asm  B  +  cos  J.  cos  B 

_  —  cos  (^  +  B) 
~    cos{A-B)  ' 
.-.  tan^  J  c  =  -  cos  {A+B)  sec  {A—B). 

8.  Deduce  from  [37]-[42]  the  for- 
mula 

tan2  J  J.  =  tan  [J  ( J.  +  ^)  -  45°] 

tan  [J  (^-5) +  45°]. 
From  [40] 

cos  A 


cos  a  =  cos  A  CSC  5  = 


sin  B 


TEACHERS     EDITION. 


121 


whence,  by  proceeding  as  before, 

1  —  cos  a  _  sin  B  —  cos  A 

1  +  cos  a     sin  B  +  cos  A 

By  [18],  page  47, 


1 


cos  a 


1  +  cos  a 


tan^  h  a. 


If  in  [6],  p.  44,  we  make  x  =  45°, 
we  have 

tan  (45°+ 2/)  = ^^ 

cosy 

_  cos  y  +  sin  y 
cos  y  —  sin  y 

And,   in    like    manner,    making 
y  =  45°  in  [10],  page  46,  we  have 


tan  (x -45°)  = 


sm  X  —  cos  X 
sin  X  +  cos  a; 


.-.  tan  (a;-45°)  tan  (45°  +  y) 

(sin  X  —  cos  x)  (sin  y  +  cos  y) 
(sin  a;  +  cos  x)  (cos  y  —  sin  y) 

sin  {x  —  y)  —  cos  (a^  +  y) 
sin  (a?  —  y)  +  cos  (cc  +  y) 

If  in  this  equation,  in  which  x 
and  y  may  have  any  values,  we 
make 

x  =  \{A^-Bl    y  =  l{A-B\ 
whence 

x-y  =  B,  x  +  y  =  A,' 

the  equation  becomes 
tan  [^(^  +  ^)- 45°] 

tan  [^  (^  -  5)  +  45°] 
_  sin  B  —  cos  A 
sin  J?  +  cos  -4 
.-.  tan2  ^  a  =  tan  [}  {A  +  B)  -  45°] 
tan  [i  {A-B)  +  45°]. 


Exercise  XXI.    Page  109. 


1.  Show  that  Napier's  Rules  lead 
to  the  equations  contained  in  For- 
mulas [38],  [39],  [40],  and  [41]. 

sin  a  =  cos  co.  c  cos  co.  A 

=  sin  c  sin  A. 
sin  h  =  cos  co.  c  cos  co.  B 

=  sin  c  sin  B. 

sin  CO.  B  =  tan  a  tan  co.  c. 
cos  5       =  tan  a  cot  c. 
sin  CO.  A  =  tan  6  tan  co.  c. 
cos  J.       =  tan  b  cot  c. 

sin  CO.  A  =  cos  a  cos  co.  ^. 
cos  A       =  cos  a  sin  B. 
sin  CO.  5  =  cos  b  cos  co.  J., 
cos  B       =  cos  6  sin  A. 


sin  a  =  tan  co.  B  tan  5 
=  cot  B  tan  6. 

sin  b  =  tan  a  tan  co.  A 
=  tan  a  cot  -4, 


2.  What  will  Napier's  Rules  be- 
come, if  we  take  as  the  five  parts  of 
the  triangle,  the  hypotenuse,  the 
two  oblique  angles,  and  the  comple- 
ments of  the  two  legs. 

(i.)  Cosine  of  middle  part  equals 
product  of  cotangents  of  adjacent 
parts. 

(ii.)  Cosine  of  middle  part  equals 
product  of  sines  of  opposite  parts. 


122 


TRIGONOMETRY. 


Exercise  XXII.    Page  114. 


1.  Solve  the  right  triangle,  given 
a  =36°  27^  5  =  43°  32^  ?>V\ 

Taking  a  as  the  middle  part,  we 
have,  hy  Eule  I., 

sin  a     =  tan  h  cot  B, 
whence       tan  h    —  sin  a  tan  B, 

■3  ^       -r,       tan  h 

and 


sin  a 

log  tan  h 

=    9.97789 

colog  sin  a 

=    0.22613 

log  tan  B   =  10.20402 

B  =  57°  59^  19.3^^ 

Taking  h  as  the  middle  part,  hy 
Rule  I., 

sin  h      =  tan  a  cot  A. 

tan  a     =  sin  &  tan  A. 
tan  a 


tan  A 


sin  h 


log  tan  a  =    9.86842 

colog  sin  h  =    0.16185 

log  tan  J.  =10.03027 

A  =  46°  59^  43.2^^. 

Taking  co.  c  as  the  middle  part, 
by  Rule  II., 

cos  c    =  COS  a  cos  h. 

log  cos  a     =  9.90546 
log  cos  h     =  9.86026 

log  cos  c     =  9.76572 
c  =  54°  20^ 

2=   Solve  the  right  triangle,  given 
a  =  ^Q°  40^  h  =  32°  40^ 


cos  c  =  COS  a 


cos  h. 


log  cos  a  =  8.76451 

log  cos  h  =  9.92522 

log  cos  c  =  8.68973 

c  =  87°  11^  39.8^^. 

tan  A  =  tan  a  esc  5. 

log  tan  a  =11.23475 

log  CSC  h  =    0.26781 

log  tan  ^  =11.50256 

A  =  88°  11^  57.8^^ 

tan  B  =  tan  h  esc  a. 

log  tan  h  =  9.80697 

log  CSC  a  =  0.00074 

log  tan  5  =9.80771 

B  =  32°  42^  38.7^^ 

3.   Solve  the  right  triangle,  given 

=  50°,  b  =  36°  54^  49^^ 

cos  c  =  COS  a  cos  b. 

tan  A  =  tan  a  esc  b. 

tan  B  =  tan  5  esc  a. 

log  cos  a  =  9.80807 

log  cos  b  =  9.90284 

log  cose  =9.71091 

c  =  59°  4^  25. 7^^ 

log  tan  a  =10.07619 

log  CSC  b  =    0.22141 

log  tan  ^  =10.29760 

A  =63°  15M3.13^^ 

log  tan  b  =  9.87575 

log  CSC  a  =  0.11575 

log  tan  5  =9.99150 

B  =  44°  26^  21 .6^^ 


TEACHEES     EDITION. 


123 


4.  Solve  the  right  triangle,  given 
a  =  120°  10^  b  =  150°  59^  U^\ 

cos  c  =  COS  a  cos  h. 
t&nA  =  tana  esc  b. 
tan  B  =  tan  b  esc  a. 

log  cos  a  =9.70115 
log  cos  &    =  9.91180 

log  cos  c    =  9.64295 

c    =  63°  55'  43.3'^ 

log  tan  a  =  10.23565 
log  esc  6    =    0.31437 

log  tan  A  =  10.55002 

J..=  105°44'21.25''. 

log  tan  5  =9.74383 
log  esc  a  =  0.06320 
log  tan  B  =  9.80703 

5  =  147°19'47.14'^ 

5.  Solve  the  right  triangle,  given 
c  =  55°  9'  32'^  a  =  22°  15'  7''. 

cos  B  =  tan  a  cot  c. 
log  tana  =9.61188 
log  cote    =  9.84266 

log  cos  B  =  9.45454 

B  =  73°  27'  11.16". 

tan  h   =  sin  a  tan  B. 
log  sin  a    =    9.57828 
log  tan  B  =  10.52709 

log  tan  6   =10.10537 
b   =51°  53'. 

cos  A  =  tan  b  cot  c. 
log  tan  b  =  10.10537 
log  cote    =    9.84266 

log  cos  A  =    9.94803 

A  -  27°  28'  25.71". 


6.  Solve  the  right  triangle,  given 
c  =  23°  49'  51",  a  =  14°  16'  35". 

cos  b  =  cos  c  sec  a. 

log  cose  =9.96130 

log  sec  a  =  0.01362 

log  cos  &  =9.97492 

b  =  19°  17'. 

sin  A  =  sin  a  esc  c. 
log  sin  a   =9.39199 

log  CSC  e    =  0.39358 
^  — —  « 

log  sin  A  =9.78557 

A  =  37°  36'  49.4'^ 

cos  B  =  tan  a  cot  c. 
log  tan  a   =    9.40562 
log  cote    =  10.33488 

log  cos  B  =   9.76050 

B  =  54°  49'  23.3". 

7.  Solve  the  right  triangle,  given 
c  =  44°  33'  17",  a  =  32°  9'  17". 

cos  b   =  cos  c  sec  a. 

log  cose    =9.85283 
log  sec  a    =  0.07231 

log  cos  b    =  9.92514 
b    =32°  41'. 

sin  A  =  sin  a  esc  c. 

log  sin  a    =9.72608 
log  CSC  e    =  0.15391 

log  sin  A  =  9.87999 

A  =  49°  20'  16.3". 

COS  B  =  tan  a  cot  c. 

log  tan  a  =  9.79840 
log  cot  c  =  10.00675 
log  cos  B  =    9.80515 

B  =  50°  19'  16'^ 


124 


TRIGONOMETEY. 


.  8.   Solve  the  riglit  triangle,  given 
c  =  97°  13^  4^^  a  =  132°  1¥  12'\ 

cos  b   =  cos  c  sec  a. 

log  cose  =9.09914 
log  sec  a   =  0.17250 

log  cos  6    =9.27164 

h   =  79°  13^  38.18^^ 

sin  A  =  sin  a  esc  c. 

log  sin  a  =  9.86945 
log  esc  c  =  0.00345 
log  sin  A  =  9.87290 

A  =  48°  16^  10^^ 

But  A  and  a  must  be  of  the  same 
kind, 

.-.  A  =  131°  43^  50^^ 

cos  B  =  tan  a  cot  c. 
log  tan  a  =  10.04196 


log  cot  c 


9.10259 


log  cos  B  =    9.14455 

B  =  81°  58^  53.3^^. 

9.  Solve  the  right  triangle,  given 
a  =  77°  21^  50^^  A  =  83°  56^  40^^ 

sin  c  =  sin  a  esc  A. 

log  sin  a   =  9.98935 
log  CSC  ^  =0.00243 

log  sin  c    =  9.99178 

c    =  78°  53^  20^^. 

Since  c  is  found  from  its  sine,  it 
may  have  two  values  which  are 
supplements  of  each  other. 

Hence  also      c  =  101°  6^  40^^ 

sin  h  =  tan  a  cot  A. 

log  tana  =10.64939 
log  cot  A  =    9.02565 

log  sin  b   =    9.67504 


or 


b   =    28°  14^  31.3^^. 
=  151°  45^  28.7^^ 


sin  B  =  sec  a  cos  A. 

log  sec  a   =0.66004 
log  cos  A  =  9.02323 

log  sin  B  =  9.68327 

B  =    28°  49^  57.4^^, 
or  =  151°  10^    2.6^^ 

10.  Solve  the  right  triangle,  given 
a  =  77°  2V  50^^  A  =  40°  40^  40^^ 

sin  c    =  sin  a  esc  A. 

log  sin  a   =    9.98935 
log  CSC  J.  =    0.18588 


log  sin  c 


=  10.17523 
.".  sin  c  >  1,  which  is  impossible. 

11.  Solve  the  right  triangle,  given 
a  =  92°  47^  32^^  B  =  50°  V  V\ 

tan  e    =  tan  a  sec  B. 

log  tan  a  =11.31183 
log  sec  B  =    0.19223 

log  tan  c   =  11.50406 

c  =91°47M0^^ 

tan  b    =  sin  a  tan  B. 

log  sin  a  =    9.99948 
log  tan  B  =  10.07671 
log  tan  b  =  10.07619 
b  =50°. 

cos  A  =  cos  a  sin  B. 

log  cos  a  =  8.68765 
log  sin  B  =  9.88447 
log  cos  .4  =8.57212 

A  =  92°  8^  23^^. 


TEACHERS     EDITION. 


125 


12.  Solve  the  right  triangle,  given 
o  =  2°  0^  55^^  B  =  12°  40\ 

tan  h  =  sin  a  tan  B. 

log  sin  a  =8.54612 

log  tan  5  -9.35170 

log  tan  &  =  7.89782 

h  =0°  27^  10.2^^ 

tan  c  =  tan  a  sec  B. 

log  tan  a  =  8.54639 

log  sec  ^  -0.01070 

log  tan  c  =  8.55709 

c  =  2°  3'  55.7^^ 

cos  A  =  cos  a  sin  ^. 

log  cos  a  =  9.99973 

log  sin  B  =  9.34100 

log  cos  ^  =9.34073 

A  =  77°  20^  28.4^^ 

13,  Solve  the  right  triangle,  given 
a  =  20°  20^  20^^  B  =  38°  10^  10^^ 

tan  b    =  sin  a  tan  B. 


log  sin  a 

=  9.54104 

log  tan  B 

=  9.89545 

log  tan  b 

=  9.43649 

b 

=  15°  16^  50.4^^ 

tan  c 

=  tan  a  sec  B. 

log  tan  a 

=  9.56900 

log  sec  -S 

=  0.10448 

log  tan  c 

=  9.67348 

c 

=  25°  14^  38.2^^ 

cos  A 

=  cos  a  sin  B. 

log  cos  a 

=  9.97204 

log  sin  B 

=  9.79098 

log  cos  J. 

=  9.76302 

A 

=  54°  35^  16.7^^ 

14.  Solve  the  right  triangle,  given 
a  =  54°  30^  B  =  35°  30^ 

tan  c    =  tan  a  sec  B. 
log  tan  a  =  10.14673 
log  sec  B  =    0.08931 

log  tan  c    =10.23604 

c   =  59°  51^  20.7^^. 

tan  b  =  sin  a  tan  B. 
log  sin  a    =9.91069 
log  tan  B  =  9.85327 
log  tan  6   =9.76396 

b  =  30°  8^  39.3^^ 

cos  A  =  cos  a  sin  B. 

log  cos  a  =  9.76395 
log  sin  B  =  9.76395 
log  cos  A  =  9.52790 

A  =  70°  17^  35^^. 

15.  Solve  the  right  triangle,  given 
c  =  69°  25^  11^^,  A  =  54°  54^  42^^. 

sin  a    =  sin  c  sin  A. 
log  sin  c    =  9.97136 
log  sin  A  =  9.91289 
log  sin  a    =  9.88425 
a  =50°. 

tan  b    =  tan  c  cos  A. 
log  tan  c  =  10.42541 
log  cos  A  =    9.75954 

log  tan  6   =10.18495 

b  =  56°  50^  49.3^^ 

cot  B  =  cos  c  tan  A. 
log  cos  c    =    9.54595 
log  tan  A  =  10.15335 
log  cot  B  =    9.69930 

B  =  63°  25^  4^^. 


126 


TRIGONOMETRY. 


16.  Solve  the  right  triangle,  given 
c  =  112°  48^  A  =  56°  1 V  56^^ 

cot  B  =  cos  c  tan  A. 

log  cose  =  9.58829 
log  tan  J.  =  10.17427 
log  cot  ^=    9.76256 

£  =  120°  3^  50^^ 

sin  a  =  sin  c  sin  A. 

log  sin  c    =  9.96467 
log  sin  ^  =  9.91958 

log  sin  a   =9.88425 
a  =  50°. 

tan  b  =  cos  A  tan  c. 

log  cos  A  =  9.74532 
log  tan  c  =10.37638 

log  tan  5  =10.12170 

b  =  127°    4^  30^^ 


17.  Solve  the  ri^ 

^ht  triangle,  given 

c  =  46°  40^ 

12^^ 

A  =  37°  46^  d''. 

sin  a 

=  sin  A  sin  c. 

log 

sm  A 

=  9.78709 

log 

sin  G 

=  9.86178 

log 

sin  a 

=  9.64887 

a 

=  26°  27^  24^^ 

tan  b 

=  tan  c  cos  A. 

log 

tanc 

=  10.02533 

log 

cos  A 

=    9.89789 

log 

tan  b 

=    9.92322 

b 

=  39°  57^  41.5^^ 

cot^ 

=  tan  A  cos  c. 

log 

cos  c 

=  9.83645 

log 

tan  J. 

=  9.88920 

log 

cot  5 

=  9.72565 

£ 

=  62°  0^  4'^ 

18.  Solve  the  right  triangle,  given 
c  =  118°  40^  V^,  A  =  128°  0^  4^^ 

sin  a  =  sin  c  sin  A. 

log  sin  G  =  9.94321 
log  sin  A  =  9.89652 

log  sin  a  =  9.83973 

a  =  136°  15^  32.3^^. 

tan  b  =  tan  c  cos  A. 

log  tanc  =10.26222 
log  cos  J.  =    9.78935 

log  tan  6  =  10.05157 

b  =  48°  23^  38.4^^. 

cot  B  =  cos  c  tan  A. 

log  cos  c  =  9.68098 
log  tan  J.  =  1010717 
log  cot  5=    9.78815 

B  =  58°  27^  4.3^^. 

19.  Solve  the  right  triangle,  given 
^  =  63°15^12^^  ^=135°  33^  39^^. 

cos  c  =  cot  A  cot  B. 
log  cot  ^=    9.70241 
log  cot  B  =  10.00850 

log  cos  c  =    9.71091 

c  =  120°  55^  34.3^^. 

cos  a  =  cos  A  CSC  B. 

log  cos  A  =  9.65326 
colog  sm  B  =  0.15480 

log  cos  a  =  9.80806 

a  =  49°  59^  56^^ 

cos  b  =  cos  B  CSC  A. 

log  COS  .5  =  9.85369 
colog  sin  A  =  0.04915 

log  COS  b  =-9.90284 

b  =  143°  5^  12^^ 


TEACHERS     EDITION. 


127 


20.  Solve  the  right  triangle,  given 
A  =  116°  43^  12^^  B  =  116°  3V  25'\ 

cos  a  =  cos  A  CSC  £. 

log  cos  A  =  9.65286 
log  CSC  B  =  0.04830 

log  cos  a  =  9.70116 

a  =120°  10^  3^^ 

cos  b  =  cos  -B  CSC  A. 

log  cos  5  =  9.64988 
log  CSC  J.  =  0.04904 

log  cos  h  =  9.69892 

b  =119°59M6^^. 

log  cos  c  =  cot  A  cot  B. 

log  cot  ^  =  9.70190 
log  cot  5  =  9.69818 

log  cos  c  =  9.40008 

c  =  75°  26^  58^^ 

21.  Solve  the  right  triangle,  given 
A  =  46°  59^  42^^,  B  =  57°  59^  17^^. 

cos  a  =  COS  A  CSC  B. 

log  cos  A  =  9.83382 

log  esc  B  =  0.07164 

log  cos  a  =  9.90546 

a  =  36°  27^ 

cos  b  =  cos  B  CSC  A. 

log  cos  B  =  9.72435 
log  CSC  A  =  0.13591 
log  cos  b  =  9.86026 

b  =  43°  32^  37^^ 

cos  c  =  cot  A  cot  B. 

log  cot  ^  =  9.96973 

log  cot  B  =  9.79599 

log  cos  c  =  9.76572 

c  =  54°  20^ 


22.  Solve  the  right  triangle,  given 
A  =  90°,  B  =  88°  24^  35^^. 

cos  c  =  cot  A  cot  B. 

But  cot  J.  =  0. 

.•.  cos  c   =  0. 

c   =90°. 

cos  a  =  cos  J.  CSC  B. 

But  COS  .A  =  0. 

.•.  cos  a  =  0. 
a  =90°. 

cos  b  =  cos  B  CSC  J.. 

CSC  A  =  l. 
b  =B. 
b  =  88°  24^  35^^. 

23.  Define  a  quadrantal  triangle, 
and  show  how  its  solution  may  be 
reduced  to  that  of  the  right  triangle. 

A  quadrantal  triangle  is  a  tri- 
angle having  one  or  more  of  its 
sides  equal  to  a  quadrant. 

Let  A^B'Q^  be  a  quadrantal  tri- 
angle with  side  A'B'=  90°,  or  a 
quadrant. 

Let  ABQ\)^  its  polar  triangle. 

Then  since 

A'B'^  (7=180°,  C=90°. 

.-.  ABQ\%  a  right  triangle. 

.•.  all  parts  of  the  polar  triangle 
may  be  found  by  formulas  for  right 
triangle.  The  parts  of  A^B^C^  may 
then  be  found  by  subtracting  proper 
parts  of  A^Cfrom  180°. 

24.  Solve  the  quadrantal  triangle 
whose  sides  are  : 

a  =  174°  12^  49.1^^ 
b=   94°    8^20'^, 
c  =    90°. 


128 


TRIGONOMETRY. 


Let  A^,  B^,  C,  a\  ¥,  c'  repre- 
sent the  corresponding  angles  and 
sides  of  the  polar  triangle. 

Then  A'^    5=>  47^  10.9^^, 
B'=  85°  bV  W\ 
C^=  90°. 

tan^  \  c' 

=  -  cos  {B'+A')  sec  {B^-A'). 

B'+  A^=  91°  38^  50.9^^ 

B^-  A^=  80°    4^  29.  F^. 

log  cos  (5^+ J-O    =8.45863 

log  sec  (^^- .40    =  0.76356 

2)9.22219 

log  tan  Jc^  =  9.61110 

Jc^=    22°  12^561^^. 
c^  =   44°  25^  53^^. 
C==135°34^    7^^ 

tan2  J  b'=  tan  [J  {B'+  A^  -  45°] 

tan[45°+H^^-^0]- 
J  (^^+  .SO  -  45°  =  49^  25.5^^ 
45°  +  i  {B'-  A^)  =  85°  2^  14.6^^ 

log  tan   0°  49^  25.5^^=    8.15770 

log  tan  85°   2^4.6^^=11.06133 

2)  9.21903 

log  tan  1 6^=    9.60952 

J&^=  22°  8^35^^ 
6^=  44°  17^  10^^. 
5  =135°  42^  50^^ 

tan2  J a'=  tan  [J  (^^+  ^0  -  45°] 

tan[45°-H^'-^0]- 
H-S^+^0  -  45°  =  0°  49^  25.5^^. 
45°  -  J  {B'-A')  =  4°  57^  45.4^^ 


log  tan  0°  49^  25.5^^=  8.15770 

log  tan  4°  57^  45.4^/=  8.93867 

2)  7.09637 

log  tan  ^a^=  8.54819 

ia^=  2°  1^25^''. 
a^=  4°  2^50^^. 
^  =  175°57^10^^ 


25.  Solve  the  quadrantal  triangle 
in  which 

c  =  90°, 

A  =  110°  47^  50^^ 

5=135°35^34.5^^ 

Let  A\  B\  (7,  a',  ¥,  c^  repre- 
sent the  corresponding  angles  and 
sides  of  the  polar  triangle. 

Then    a'=  69°  12^  10^^. 
&/=  44°  24^  25.5^^ 
(7=  90°. 

tan  A^=  tan  a^  esc  h^. 

log  tan  a'  =  10.42043 
log  CSC  y  =  0.15505 
log  tan  A^=  10.57548 

A^=    75°    6^58^^ 
a  =  104°  53^    2^^. 

tan  B^=  tan  h^  csc  a^. 

log  tan  5^=  9.99101 
log  CSC  a^  =  0.02926 
log  tan  B^=  10.02027 

B^=   46°20M2^^ 
b  =  133°  39^  48^^ 

cos  c'  =  cot  J.''  cot  B^. 


TEACHERS     EDITION. 


129 


26. 

A,  Q 
angle. 


log  cot  A'=^  9.42452 
log  cot  B^=  9.97973 
log  cos  c^  =  9.40425 

c/  =    75°  18^  2V^. 
(7=104°4F39^^ 

Given  in  a  spherical  triangle 
and  c  =  90°  ;    solve   the   tri- 

sin  a  =  sin  c  sin  A. 

=  1x1. 

.'.  a  =  90°. 
tan  b  =  tan  c  cos  A 
=  cox0. 
.-.  b  =  45°. 
cot  5  =  cos  c  tan  A. 
=  0xco. 
.-.  B  =  45°, 


27.   Given  ^  =  60°,  C=  90°,  and 
c  =  90° ;  solve  the  triangle. 

sin  a  =  sin  c  sin  A. 

tan  b  =  tan  c  cos  yl. 

cot  5  =  cos  c  tan  J.. 

sin   c  =  1. 

.•.  sin   a  =  sin  A. 

a  =  A  =  60°. 

tan  c  =  00. 
.".  tan  5  =  00. 
6  =  90°. 
cos  e  =  0. 
.'•.cot^  =  0. 
B  =  90°. 


28.  Given  in  a  right  spherical 
triangle,  A  =  42°  24^  9^^  ^  =  9°  4^ 
W^  ;  solve  the  triangle. 

cos  c   =  cot  J.  cot  B. 

log  cot  tI  =  10.03943 
log  cot  B  =  10.79688 
log  cos  c  =  10.83631 
which  is  impossible. 

.".  triangle  is  impossible. 

29.  In  a  right  triangle,  given 
a  =  119°  IF,  ^  =  126°54^  solve 
the  triangle. 

tan  c  =  tan  a  sec  B. 

log  tan  a  =  10.25298 
log  cos  B  =    0.22154 

log  tan  c  =  10.47452 

c  =  71°  27^  43^^ 

tan  b  =  sin  a  tan  5. 

log  sin  a  =  9.94105 
log  tan  B  =  10.12446 
log  tan  b  =  10.06551 

6  =130°4F42^^ 

,  cos  .A  =  cos  a  sin  5. 

log  cos  a  =9.68807 
log  sin  B  =  9.90292 

log  cos  A  =  9.59099 

^  =  112°  57^  2^\ 

30.  In  a  right  triangle,  given 
c  =  50°,  b  =  44°  18^  39^^ ;  solve  the 
triangle. 

cos  a  =  cos  c  sec  b. 

log  cos  e  =  9.80807 
colog  cos  b  =  0.14535 

log  cos  a  =  9.95342 

a  =  26°  3^  51^^. 


130 


TRIGONOMETRY. 


sin  A  = 

sin  a  CSC  c. 

log  sin  a  = 

9.64284 

log  CSC  c    = 

0.11575 

log  sin  A  = 

9.75859 

A  = 

35°. 

tan  5  = 

tan  h  CSC  a. 

log  tan  b  = 

9.9S955 

log  CSC  a  = 

0.35716 

log  tan  B  = 

10.34671 

£^ 

65°  46^  7'' 

31.  In  a  right  triangle,  given 
A  =  156°  20'  30^^  a  =  65°  15^45^^; 
solve  the  triangle. 

It  is  impossible,  because  a  and  A 
are  unlike  in  kind.  And,  in  Case 
III.,  they  must  be  alike  in  kind; 
otherwise,  impossible. 

32.  If  the  legs  a  and  5  of  a  right 
spherical  triangle  are  equal,  prove 


that  cos  a 

=  cot  J.  =Vcosc. 

• 

COS  c  =  COS  a  cos  b. 

But 

cos  a  =  cos  b. 

•,  COS  c  =  cos^a. 

and 

•.  cos  a  =  Vcos  c. 

sin  a  =  cos  a  sin  b  tan  J.. 

Since 

sin  a  =  sin  5, 

1  =  cos  a  tan  A. 

cos  a  = 

tan  A 

.  cos  a  =  cot  A. 

33.  In  a  right  triangle  prove  that 
co&^A  X  sin^c  =  sin  (c  —  a)  sin  (c  +  a). 

cos  J.  sin  c  =  cos  a  sin  b. 

.".  cos^^  sin^c  =  cos^a  sin^b.    (1) 


sin^6  =  sin^c  sin^^ 

=  sin^c  (1-C0S2.5).    (2) 
cos^B  =  tan^a  cot^c 


sin^a 
cos^a 

cos^c 

-^    ■   9  ■ 

sm-^c 

Substitute 

m  (2), 

sin^  b  = 

sin^?  — 

sin^a  cos^c 
cos'^^a 

sin^c  cos^a— sin^a  cos^c 

cos'^'a 
Substitute  in  (1), 
cos^a  sin^Z)  =  sin^c  coshes  — sin^a  cos^c 
=  (sin  c  cos  a  +  sin  a  cos  c) 
(sin  c  cos  a  —  sin  a  cos  c) 
=  sin  (c  +  a)  sin  (c  —  a). 
Substitute  in  (1), 
cos^^  sin^c  =  sin  (c  +  a)  sin  (c  —  a). 

34.   In  a  right  triangle  prove  that 
tan  a  cos  c  =  sin  b  cot  ^. 

sin  &  =  tan  a  cot  -4. 

,    >      sin  & 
cot -4=  ■ ^• 

tan  a 

cos  c  =  cot  -4  cot  B. 


cot  A 

cos  c 
cotB 

cos  c 

sin  6 

cot  B     tan  a 
tan  a  cos  c  ==  sin  6  cot  ^. 

35.   In  a  right  triangle  prove  that 
sin^-4  =  cos^^  +  sin%  sin^B. 

sin  o  =  sin  ^  sin  c. 
•  o  A      sin^a 
sin^c 
cos  B  =  tan  a  cot  e. 
cos^5  =  tan^a  cot^c. 


TEACHERS     EDITION. 


131 


sinM—  cos'^B 


:  — tan^a  cot^'c 

sin^a     sin^a      cos^c 
sin^c      cos^a     sin^c 

sin%  cos^a  —  sin^a  cos^c 


cos'^a  sm^c 


_  sin%  (cos^a  —  cos^c) 


Now 


cos^'a  sm'^c 


cos  c  =  cos  a  cos  6. 


(1) 


cos^a  = 


cos^c 


cos^S 
cos^c  =  cos^a  cos^5. 
sin  b 


sm  c  = 


sm^c  = 


sin^ 
sin^Z) 


Substitute  these  values  in  (1), 
sin^a  —  co&'^B 


=  sin'^a 


'cos-'c 
cos^6 


cos^a  cos^S 


cot^^c      sin-6 
X 


=  sin^a 


=  sin^a 


_     cos'-^i     sin^^ 

'"cos^c  —  cos^g  cos^5 

cos^6 

cos^c  sin^6 


sin^i? 


cos^6  sin^5 

cos'^c— cos^acos*& 


cos^c  sin^6 


•  2     •  9  D /cos^c  — cos'^acos^SX 

sin^a  sm^ij  f  : ) 

cos^c  sin^6       J 


But 


cos^c  —  cos%  cos*6 
cos^c  sin^5 


-Q^2  _  ODx  OE^- 

0~D' 

0E\  DE' 

0E\  OD^  -  OE^ 
0E\  ED'' 

on^~OE^ 


ED" 


=  1. 


.•.  sin^^  —  cos^^  =  sin^a  sin^^. 
.•.  sinM  =  cos^^  +  sin^a  sin^^. 

36.  In  a  right  triangle  prove  that 
sin  (6  +  c)  =  2  cos^  J  a  cos  6  sin  c. 

sin  (b  +  g) 

=  sin  5  cos  c  +  cos  h  sin  c 

sin  &  cos  c  ,  T  \       7    • 

; — : —  +  1   cos  0  sm  c 

cos  0  sm  c        J 

=  (tan  b  cot  c+1)  cos  b  sin  c.  (1) 

But  tan  b  cot  c  =  cos  A, 

.•.  tanJ  cote +1  ^  cos  J.  +  1.      (2) 


132 


TRIGONOMETRY. 


4 


cos  ^  +  1 


COS  hA. 


Substitute  in  (2), 
(tan  bc^e  +  1)  =  2  cos^  J  A. 

Substitute  in  (1), 

sin  (6  +  c)  =  2  cos^  ^A  cos  h  sin  c. 

37.  In  a  right  triangle  prove  that 
sin  {c  —  b)  =  2  sin^  ^a  cos  6  sin  c. 

sin  (c  —  b) 

=  sin  c  cos  b  —  cos  c  sin  b 

7  A-,      cosf  sin  b\  ^^s 
=  sm  c  cos  0    1  — : (1) 

\        sin  c  cos  oy 
=  sin  c  cos  6(1— cote  tan  6).     (2) 
cot  c  tan  b  =  cos  A. 

1  —  cos  A 


1— cote  tan  5  =  2 


(3) 


sinM  =  '\~ 


cos  A 


Substitute  in  (3), 
1  —  cot  c  tan  5  =  2  sin^  |  A. 

Substitute  in  (2), 

sin  (c  —  5)  =  2  sin^  J  a  sin  e  cos  b. 

38.  If,  in  a  right  triangle,  p  de- 
note the  arc  of  the  great  circle  pass- 
ing through  the  vertex  of  the  right 
angle  and  perpendicular  to  the 
hypotenuse,  w  and  n  the  segments 
of  the  hypotenuse  made  by  this  arc 
adjacent  to  the  legs  a  and  b  respec- 
tively, prove  that 

(i.)  tan^a  =  tan  c  tan  m, 

(ii.)  sin^p  =  tan  m  tan  n. 


In  triangle  BCA 

cos  B  =  tan  a  cot  c. 

,  cos  B 

:.  tan  a  = 

cot  c 

In  triangle  CBD 

tan  5(7=  tan  5Z)  sec  5. 

tan  a     =  tan  ni  sec  B 

_  tan  m 
cos  5 

Multiplying  the  two  equations, 

,      o        tan  m  ^,  cos  5 

tan''a  = X 

cos  B      cot  c 

=  tan  m  tan  c. 

2d.  In  triangle  CBD 

sinp  =  tan  m  cot  ili"; 
and  in  triangle  CAD 

sinp  =  tan  n  cot  iV! 
But  cot  Mx  cot  i\^=  1. 
.-.  if+iV"=90°. 

sin^p  =  tan  m  tan  w. 


TEACHERS     EDITION. 


133 


Exercise  XXIII.    Page  116. 


1.  In  an  isosceles  spherical  tri- 
angle, given  the  base  h  and  the  side 
a  ;  find  A  the  angle  at  the  base,  B 
the  angle  at  the  vertex,  and  h  the 
altitude. 

Let  ABA^  be  an  isosceles  trian- 
gle, A  and  A^  being  the  equal 
angles,  a  and  a^  the  equal  sides. 

Let  h  the  arc  of  a  great  circle  be 
drawn  from  B  perpendicular  to 
AA'. 

Let  h  and  co.  c  be  the  given  parts. 
6  =  J  6   in  triangle  ABA\ 
c  =    a  in  triangle  ABA\ 
B  =  ^B  in  triangle  ABA\ 

cos  A    =  cot  a  tan  |  b. 
sin  2"  &  =  sin  a  sin  ^  B. 
sin^  B  =  esc  a  sin  |  b. 
cos  A     =  cos  a  sec  J  &. 

2.  In  an  equilateral  spherical 
triangle,  given  the  side  a  ;  find  the 
angle  A. 

In  the  equilat.  triangle  AA^A^^ 
draw  &rc  AC ±  to  A^A^^. 

Then  in  right  triangle  AA^C, 
given  a. 

sin  ^  J.  =  sin  J  a  esc  a 


=V^ 


cos  a        1 


sma 


/    1  —  cos  a 
^'2  (1  -  cos^a) 

'2(1 -(-COS  a) 

2   ^  1  4-  cos  a 
^  sec  I  a. 


Also,  cos  -4  =  tan  ^  a  cot  a 


V 


1— cosa       cos% 


l-hcosa     1— cos^a 

=  cos  a  X 

1  -I-  cos  a 

=  cos  a  J  sec^  J  a. 


3.  Given  the  side  a  of  a  regular 
spherical  polygon  of  n  sides  ;  find 
the  angle  A  of  the  polygon,  the 
distance  H  .from  the  centre  of  the 
polygon  to  one  of  its  vertices,  and 
the  distance  r  from  the  centre  to  the 
middle  point  of  one  of  its  sides. 


In  the  regular  polygon  ABDE 
draw  arcs  from  the  vertices  A,  B, 
etc.,  through  the  centre  C,  and  from 
Cto  M,  the  middle  of  one  side. 


134 


TRIGONOMETRY. 


Then 

n 

also 

n 

and 

CAM  =  I  A, 

AM=\a, 

AC=B, 

and 

MO=r. 

In  equilateral  right  triangle  CMA 
represent  the  parts  by  letters  A^, 
B\  C\  etc. 

sin  A^  =  cos  B'  sec  V. 

Substituting  known  values, 

sin  ^  ^  =  sec  \  a  cos 

n 

Or,         sin  c'     =  sin  h^  esc  B^, 
sm  li    =  sm  ^  a  esc 


180° 


Or,        sin  a'    =  tan  b^  cot  B^, 

180^ 


sin  r      =  tan  i  a  cot 


4.  Compute  the  dihedral  angles 
made  by  the  faces  of  the  five  regu- 
lar polyhedrons. 

Let  the  figure  ABCD  represent  a 
tetrahedron. 

It  is  required  to  find  the  dihedral 
angle  ADCB  or  the  corresponding 
plane  angle  DFO. 

DAF=  60° 

(since  it  is  an  angle  of  an  equilat- 
eral triangle). 

DFA  =  rt.  triangle 

(DF  being  JL  to  AC,  since  it  is  a 
side  of  the  plane  angle  DFO  corre- 
sponding to  the  dihedral  angle 
ABCB). 


Let 

AD  = 

1, 

Then 

DF 
AD 

sin  60°. 

DF  = 

AD  sin  ( 

30°. 

log 

AD      = 

0.00000 

log 

sin  60°  = 

=  9.93753 

-10 

log 

DF      = 

AF 
AD 

=  9.93753 
=  cos  60°. 

-10 

AF=^ 

=  AD  cos 

60°. 

log 

AD      = 

=  0.00000 

log 

cos  60°  = 

=  9.69897 

-10 

log  AF      =  9.69897  -  10 

Draw  OA  from  centre  of  triangle 
ABC. 

OAF  =  I  CAB  =2,0° 

(since  OA  produced  to  the  middle 
of  the  opposite  side  CB  will  pass 
through  the  centre  0  and  will  also 
bisect  the  angle  CAB,  CA  and  AB 
being  equal), 

OF  A  =  rt.  triangle 
(since  Oi^  is  ±  to  ^C). 


OF 


=  tan  30°. 


AF 

Oi^=Ai^  tan  30°. 


log  AF      =  9.69897 
log  tan  30°=  9.76144 

log  OF 


10 


OF 


9.46041  - 10 


=  cos  DFO. 


DF 

log  OF      =  9.46041  - 10 
colog  DF      =  0.06247 
log  cos  DFO  =  9.52288  -  10 
DFO  =  70°  31^  43^^ 


TEACHEES     EDITION. 


135 


Let  AC^  be  a  cube.  E,equired 
the  dihedral  angle  AB^D^. 

The  lines  AA^  and  BB^  deter- 
mine the  plane  ABB^A\  and  the 
lines  A^D^  and  B^C^  determine  the 
plane  A^D^C^B^-  and  as  AA^  and 
BB'  are  ±  to  A' B'  and  B'C\  re- 
spectively, the  planes  must  be  per- 
pendicular. 

.".  the  angle  required  is  90°. 

Let  E-ABCD-F  be  an  octahe- 
dron. Required  the  dihedral  angle 
E-BA-F, 


EKO  =    54°  44^    9^^. 
EKF=^  2  EKO  =  109°  28^  18^^. 


Draw  ^^and  FK 1.  to  AB,  also 
0.2"  from  intersection  of  axes.  Then 
is  EKF  the  plane  angle  required. 

Let  AB=\, 

Then  0K=  0.5 

(since  0^,  OB,  and  0^  are  perpen- 
dicular and  equal), 

and     0E{=  OA)  =  sin  45° 

=  0.7071. 

cot  EKO  =  ^• 
OE 

log  OiT  -  9.69897 

colog  OE    =0.15051 

log  cot  EKO  =  9.84948 


Let  AE  and  DE  be  two  faces  of 

a  dodecagon. 

It  is  required  to  find  the  dihedral 

angle  AOD. 

B 

y  '^  >^    \ 

/ 

Take  each  side  of  the  pentagons 
^^andi)^=2. 

•  5  =  108°,  and  A  =  36°. 
log  AC  =  log  2  -h  log  sin  108° 
+  colog  sm  36° 

log  2  =  0.40103 

log  sin  108°   =9.97821 

colog  sin  36°     =  0.23078 

log  AC  =  0.51002 

^C=3.286L 

^C0  =  72°. 
log  AC  =  0.51002 

log  sin  72°     =  9.97821 

log  AO  =  0.48823 

^0=3.0777. 

CB  =  AC  and  AO  =  DO. 

ACD  =  108°,  being  an  angle  of  a 
regular  pentagon. 

CD  A  =  36°. 
log  AD  =  log  AC  +  log  sin  108° 
+  colog  sin  36°. 


136 


TRIGONOMETEY. 


log  AC  =  0.51002 

log  sin  108°   =9.97821 

colog  sin  36°     =  0.23078 

log  AD  =  0.71901 

^Z)  =  5.2361. 


J  ^Z)  =  2.61805. 


log  sin  J  J.  Oi)  =  log  IAD + colog  J.  0. 

log  I  AD       =0.41798 

colog  AO  =  9.51177 

log  sin  |AOi)=  9.92975 

IA0D=    58°16^52^^ 
AC>i)=116°33M4^^ 


Let  AOB,  BOC,  COD,  etc.,  be 
equilateral  triangles  forming  five  of 
the  surfaces  of  a  regular  icosahe- 
dron,  and  let  AB,  BC,  CD,  etc.  =  1. 

Kegarding  ABCDE  as  a  plane 
pentagon,  each  angle  =  108°. 

.-.  A5C=108°, 

BAC=2>Q°. 

In  a  triangle  of  which  sides  are 
AB,  BC,  and  ADC  (regarding  ADC 
as  a  straight  line  joining  centres  of 
bases  of  triangles  AOB  and  BOC), 

sin  DCB  :AB::  sin  ABC:  ADC 


log  AB  =  0.00000 

log  sin  ABC  =  9.93753 

colog  sin  DCB  =  0.30103 

log  ADC  ' 


But 


=  0.23856 
ADC  =  1.73204:. 
AD=^iADC 

=  0.86602. 


AC  is  a  diagonal  bf  plane  penta- 
gon ABCDE. 

.-.  sin  FCB  -.AB::  sin  ABC:  AC. 


AC^ 


AB^mABC 


.-.ADC^ 


AB  sin  ABC 
s'm  DCB    ' 


sin  FCB 

log  AB  =  0.00000 

log  sin  ADC=  9.97821 

colog  sin  FBC  =  0.23078 

log  AC-  =  0.20899 

AC=  1.61804. 

But  AF=^AC 

=  0.80902. 

In  right  triangle  AFD 

sin  ADF=  — , 
AD 

or  log  sin  ADF=  log  Ai^+colog  AD. 

log  Ai^  =  9.90796 

colog  AD  =  0.06247 

log  sin  ADF=  9.97043 

ADi^=  69°  5^  48^^ 

But  ADi^=JA-OD-C. 

.-.  A-OB~C=  138°  IF  36^^ 

5.  A  spherical  square  is  a  spheri- 
cal quadrilateral  which  has  equal 
sides  and  equal  angles.  Its  two 
diagonals  divide  it  into  four  equal 
right  triangles.  Find  the  angle  A 
of  the  square,  having  given  the 
side  a. 


TEACHERS     EDITION. 


137 


B 


'""fl 

^ 

y  /  \ 

\ 

\^ 

z'  /  / 

\ 

\^ 

^^V   \ 

\ 

XcB 

X     ~~^-~    1 

\ 

\ 

/  \             /"-- 

--*: 

,               \ 

T? 

"■■---A 

^\\ 

H 

yh 

Let  ABCX  be  a  spherical  square, 
and  CA,  BX  the  two  diagonals; 
also,  let  the  side  BA  =  x,  then  XA 
will  =  X  (being  equal  sides). 

By  [37],    cos  a  =  cos  x  X  cos  x 
=  cos'^o;. 


Also 


cos  X  =  Vcos  a. 
BE 


sm  X  = 


tana;  = 


OD 

BD 
OD 


Dividing   first    equation   by   the 
second, 

DE        ,  ^ 

cos  X  =  -zr^  =  cot  X. 


But 


BD 
X^iA 


(diagonals   of  a  square  bisect  the 
angles). 

Substitute  cot  J  A  for  cos  x  ; 
cot  i  A=  Vcos  a. 


Exercise  XXIV.    Page  119. 


1.  What  do  Formulas  [43]  be- 
come if  ^=90°?  ifj5  =  90°?  if 
(7=90°?  if  a  =  90°?  \fA  =  B  = 
90°?   ifa  =  &  =  90°? 

If  ^  =  90°, 

sin  a  sin  ^  =  sin  h, 

sin  a  sin  C  =  sin  c. 
If  5  =  90°, 

sin  a  =  sin  5  sin  A, 

sin  6  sin  C  =  sin  c. 
If  C=  90°, 

sin  a  =  sin  c  sin  A, 

sin  b  =  sin  c  sin  B. 
If  a  =  90°, 

sin  B  =  8'mb  sin  A, 

sin  C  =  sin  c  sin  A. 

If  A^B  =  90°, 
sin  a  =  sin  b. 

If  a  =  b  =90°, 
sin  B  =  sin  A. 


2.  What  does  the  first  of  [44]  be- 
come if  ^  =  0°?  if  A  =  90°?  if 
A  =  180°  ? 

If^  =  0°, 

cos  a  =  cos  5  cos  c  +  sin  5  sin  c. 
If  ^  =  90°, 

cos  a  =  cos  b  cos  c. 
If  ^  =  180°, 

cos  a  =  cos  b  cos  c  —  sin  b  sin  c. 

3.  From  Formulas  [44]  deduce 
Formulas  [45],  by  means  of  the 
relations  between  polar  triangles 
(I  45). 

Substituting  in  Formulas  [44]  for 
a,  b,  and  c,  their  equals,  180°— J.^, 
180°  -  B',  180°  -  C',  we  obtain 
cos  (180°-^^) 

=  cos  (180°  -  B^)  cos  (180°  -  (7) 
+  sin  (180°  -  B^)  sin  (180°  -  CO 
cos  (180°  -  A^). 


138 


TRIGONOMETRY. 


Substituting  for  cos  (180° -^0- 
etc.,  their  equals  —  cos  J.',  etc.,  we 
obtain 

—  cos  A^=  cos  B^  cos  C^ 

—  sin  B^  sin  C^  cos  a^. 

Multiply  by  - 1, 


cos  A^=  —  cos  B^  cos  C^ 

4-  sin  B^  sin  C^  cos  a^; 
and  similarly, 

cos  B^=  ~  cos  J.''  cos  (7 

+  sin  J.''  sin  C^  cos  5''. 
cos  (7=  —  cos  A^  cos  5'' 

+  sin  A^  sin  .5''  cos  c^. 


Exercise  XXV.    Page  124. 


1.  Write  formulas  for  finding,  by 
Napier's  Rules,  the  side  a  when  h, 
c,  and  A  are  given,  and  for  finding 
the  side  b  when  a,  c,  and  B  are 
given. 

(i.)  By  Rule  I., 

cos  A  =  cot  6  tan  m, 
whence        tan  m  =  tan  5  cos  A. 
By  Rule  II., 

cos  a  =  cos  71  cosp, 

whence        cosp  =  cos  a  sec  n. 

cos  5  =  cos  m  cosp, 

whence        cosj9  =  cos  6  seem. 

.".  cos  a  seen  =  cos  6  seem. 

Or,  since        ?i  ={c  —  m), 

cos  a  =  cos  b  sec  m  cos  (c— m). 

(li.)  By  Rule  I., 

cos  B  =  tan  n  cot  a, 
whence         tan  n  =  tan  a  cos  -S. 
By  Rule  II., 

cos  b  =  cos  m  cosp, 

whence          cosp  =  cos  b  sec  m. 

cos  a  =  cos  n  cosp, 

whence         cos  p  =  cos  a  sec  w. 

.•.  cos  b  sec  m  =  cos  a  sec  n. 

Or,  since  m  =  {c—  n), 

cos  6  =  cos  a  sec  w  cos  (c  —  n). 


2.   Given  find 

a=  88°  12^  20^^,  ^=  Q3°15nV^, 
5  =  124°  7a7^^  ^=132°17^59^^ 
C=    50°   2^   1^^;    c=  59°  4^8^^ 

H&  -  a)  =    17°  57^  28.5^^. 

^{a  +  b)^10Q°    9M8.5^^ 

JC=    25°    F    0.5^^ 

logcosJ(&-a)  =9.97831 
log  sec  J  (a  +  5)  =0.55536 
log  cot  J  (7  =9.33100 

log  tan  J  (^  +  ^)=  0.86467 

log  sec  H^  +  ^)  =  0.86868 
log  cos  J  (a +  6)  =9.44464 
log  sin  J  (7  =  9.62622 

log  cos  ^  c  =  9.93954 

J  c  =  29°  32^  9'^ 

logsinJ(&-a)  =9.48900 
log  CSC  J  (a  +  6)  =0.01751 
log  cot  iC  =0.33100 

log  tanJ(^-^)  =  9.83751 

^{B-A)=    34°  31^  24^^ 

i{A  +  B)=   97°  46^  35^^ 

A  =    63°  15^  11^^ 

.5=132°  17^59^^ 

c  =    59°    ¥  18'^ 


TEACHERS     EDITION, 


139 


3.   Given  find 

a  =120°  55^  35^^,  J.  =  129°  58^   3^^ 

5=   88°12^20^^  B=  63°  15^  9^^ 

(7=  47°  42^   V^;  c=  55°52^40^^ 

i(a-5)=    16°2F37.5^^ 

J  (a +  5)  =  104°  33^-57.5^^ 

J  (7=    23°  51^    0.5^^ 

log  cos  ^  (a -5)  =  9.98205 
log  sec  Ha +  5)  =  0.59947 
log  cot  J  C  =    0.35448 

log  tan  1{A  +  B)=  10.93600 

J(^  +  ^)=96°36^36^^. 

log  sin  Ka-^)  =9.44976 
logcsc-^a  +  J)  =0.01419 
log  cot  ^C  =0.35448 

log  tan |(J.-^)  =  9.81843 

K^-^)  =  33°2F27^^ 
HA+^)=96°36'36^^. 

A  =  129°  58^  3^^ 
B  =    63°  15'  9^^. 

log  sec  J  (J. +  ,5)  =  0.93890 
logcosKa  +  &)  =9.40053 
log  sin  ^  C  =  9.60675 

log  cos  ^-c  =9.94618 

Jc  =  27°56^20^^. 
c  =  55°  52^  40^^. 


4.   Given  find 

5  =  63°15a2^^  .5  =  88°  12^  24^^ 

c  =  47°42'   V\  C  =  55°52'42^^ 

^  =  59°  4^25^^;  a  =  50°   F40'^ 


l{h  +  c)^ 

55°  28^  36.5^^ 

Hb-c)  = 

7°  46^  35.5^^ 

iA  = 

29°  32^  12.5^^ 

logcosi(&-c)    =    9.99599 

colog  cos  i{b  +  c)    =    0.24662 

log  cot  J  ^  =10.24671 

log  tan  i{B+C)=  10.48932 

^5  +  C)  =  72°  2^  33^^ 

logsin^S-c)     =    9.13133 

colog  sin  K&  +  c)     =    0.08413 

log  cot  ^  J.  =  10.24671 

log  tan |(^- (7)=    9.46217 

^{B-C)  =  1Q°    2' 5V'. 

iIb  +  C)  =  72°    2^33^^ 

B  =  88°  12^  24^^ 

C=  55°  52^  42^^. 

logcosH6  +  c)    =9.75338 

colog  cos  J(5+  (7)  =  0.51101 

log  sin  ^^  =9.69284 

log  cos  I  a  =  9.95723 

f  a  =  25°  0^  50^^. 
a  =  50°  V  W. 

5.   Given  find 

6=  69°25ai^^    5=   56°1F57^^ 
c  =  109°  46^9^^,    (7  =  123°2F12^^, 
A=  54°  54^  42^^;    a=   67°  13'. 

\{c-h)  =  20°  10'  34''. 
^c  +  6)  =  89°  35'  45". 
I A         =27°  27' 21". 

log  cos  ^c- 6)    =    9.97250 

colog  cos  J  (c  +  ^)    =    2.15157 

log  cot  J  ^  =10.28434 

log  tan  J  (C+ 5)  =  12.40841 

\{C\B)  =  89°  46' 34.5". 


140 


TRIGONOMETRY. 


log  sin  J  (c- 6)    =    9.53770 

colog  sin  Kc  +  ^)    =    0.00001 

log  cot  J  ^  =  10.28434 

log  tan  J  (C- .5)=    9.82205 
^{Q-B)=    33°34^37.8^^ 
C=  123°  21^  12^^ 
B=    56°1F57^^ 


logcos  J(c  +  5)   =7.84843 
colog  cos  1{C+  B)  =  2.40837 


log  sin  h  A 


lo 


g  cos  ^  a 


=  9.66376 
=  9.92056 
J  a  =  33°  36^  30^^ 
a  =  67°  13^ 


Exercise  XXVI.    Page  126. 


1.  What  are  the  formnlas  for  com- 
puting A   when   B,    C,  and  a  are 
given ;  and  for  computing  B  when 
A,  C,  and  b  are  given? 
By  Rule  I., 

cos  a  =  cot  y  cot  B, 
.'.  cot  y  =  tan  B  cos  a. 
By  Rule  II., 

cos  A  =  cosp  sin  x, 

.'.  cos  p    =  COS  A  CSC  X. 

cos  B  =  cos  JO  sin  y, 
.'.  cos  p  =  COS  B  CSC  y. 
.'.  COS  J.  CSC  a?  =  COS  B  esc  y. 
Or,  since         a;  =  C—  y, 

cos  J.  =  cos^csc?/sin(C— 2/). 

(ii.)  By  Rule  I., 

cos  b  =  cot  a;  cot  A, 
.'.  cot  X   =  tan  A  cos  6. 
By  Rule  II., 

cos  A  =  C0SJ9  sin  ar, 

.*.  cos  p    =  cos  ^  CSC  X. 

cos  ^  =  cos  jD  sin  y, 
.'.  cos  p  =  COS  ^  CSC  y. 
.'.  COS  B  cscy  =  COS  .4  esc  a;. 
Or,  since  y  =  (7—  a;, 

COS  5  =  cos  ^  esc  X  sin  (C— a;). 


2.    Given  find 

^=   26°58M6^^  a=   37°14a0^^ 

B=   39°  45^0^^,  6=121°28aO^^ 

c=154°46^48^^  C=161°22ai^^ 

i{B-A)=    6°  23^  12^^. 

^{B  +  A)  =  33°  21^  58^^. 

f  c  =  77°  23^  24^^. 

logcosi(^--^)=  9.99730 
log  sec  |(^  +  ■^)  =  0.07823 
log  tan  I  c  =10.65032 

log  tan  J  (5  + a)   =10.72585 

log  sin  i(^  +  ^y=  9.74035 
logsecJ(&-a)  =0.12972 
log  cos  J  c  =  9.33908 


log  cos  2-  0 


=  9.20915 
1  (7=  80°  41^5.4^^. 


logsin  J(5-74)=  9.04625 
log  CSC  H^  + -4)  =  0.25965 
log  tan  Jc'  =10.65032 

log  tan  J  (J -a)   =    9.95622 

1(6 -a)  =    42°    7^ 

i(b  +  a)  =    79°21M0^^ 

&  =  121°  28^  10^^ 

■  a  =    37°  14^  10^^. 

(7=  161°  22^  11^^ 


TEACHERS     EDITION. 


141 


3,    Given  find 

^  =  128°4F49^/,  a=125°4F44^^ 

5=  107°  33^  20^^  h=   82°47^34^^ 

c=  124°  12^  31^^;  (7-127°22^ 

1{A-B)=  10°  34^  14.5^^. 
1{A  +  B)  =  118°  7^  34.5^^ 
Jc  =62°    6M5.5^^. 

log  cos  J  (J.  -  5)  =    9.99257 

colog  cos  1{A-^B)=    0.32660 

log  tan  J  c  =10.27624 

log  tan  J  (a +  5)   =10.59541 

l{a+h)  =104°  14^  38.5^^. 

log  sin  i  (J. -5)=    9.26351 

colog  sin  ^{A  +  B)=    0.05457 

log  tan  ^-c  =10.27624 

logtan|(a-6)  =    9.59432 

l{a-l)  =  21° 27^  5^^ 
a  =125°41^44^^. 
h  =   82°47^34^^ 

log  sin  ^(^  +  5)  =  9.94543 

colog  cos  J  (a  -h)  =  0.03118 

log  cos  ^  c  =  9.67012 

log  cos  J  a  =  9.64673 

1(7=    63°  4P. 

C=127°22^. 

4.   Given  find 

5  =  153°  17^  &^,  5  =  152°  43^  5K^ 

C=   78°43^36^^  c=  88°  12^  21^^, 

a=   86°  15^5^^;  A==   78°15M8^^. 

i(^+C')  =  116°  0^21^^. 
i(5-C)=  37°16M5^^ 
J  a  =43°    7^37.5^^ 


log  cos  J  (5 -(7)  =  9.90074 
log  sec*  (^+(7)  =  0.35807 
log  tan  J  a  =9.97159 

log  tan  J  (5 +  c)    =0.23040 

i(6  +  c)   =  120°  28^  6/>'. 

logsin  J(5-C)  =  9.78226 
log  CSC  ^5  +  C)  =  0.04636 
log  tan  J  a  =9.97159 

log  tan  1(6 -c)   =9.80021 

lih-c)  =32°  15M5/^ 

log  sin  J  (5 +(7)  =  9.95364 
log  sec  J  (6 -c)  =0.07283 
log  cos  ^  a  =9.86322 

log  cos  J  ^  =9.88969 

J  ^  =  39°  7^  54^^. 
I  =  152°  43^  51^^. 
c  =    88°  12^  21^^ 
A  =    78°  15^  48^^ 

5.   Given  find 

J.  =  125°4F44^^,  a  =  128°31M6^^, 

(7=   82°  47^  35^^,  c  =  107°33^20^^ 

5=   52°  37^  57^^;  5=  55°47^40^^' 

H^  +  C)  =  104°  14^39.5^^. 
i(A-C)=  21°  27^  4.5^^. 
J  6  =26°  18^  58.5^^ 

log  cos  J  (^-(7)  =  9.96883 
log  sec  i{A  +  C)  =  0.60896 
log  tan  J  6  =  9.69424 

log  tan  J  (a +  c)   =0.27203 

i{a  +  c)  =118°  7^33^^ 

log  sin  J  (J. +  C)  =  9.98644 
log  sec  f  (a -c)  =0.00743 
log  cos  J  5  =9.95248 

log  cos  J  5  =9.94635 

i  ^  =  27°  53^  50^^,  ^ 


142 


TRIGONOMETRY. 


logsin  J(A-C)  =  9.563i3 
log  CSC -I  (^+  (7)  =  0.01356 
log  tan  16  =  9.69:124 

log  tan  i  (a -c)   =9.27093 


Ha  -  c)   =    10°  -ZV  IS'\ 

a  =  128°  31M6^^. 

c  =  107°  33^  20^^. 

B  =   55°  47^  40^^ 


1.   Given  find 

a=   73°49^38^^  5=116°42^30^^ 
6  =  120°  53^  35^^,     c  =  120°57^27^^ 
A=  88° 52^2^^;    C=116°47^  ¥\ 

log  sin  A  =  9.99992 

log  sin  b  =  9.93355 

log  CSC  a  =  0.01753 

log  sin  B  =  9.95100 

.5=  [180°- (63°  17^  30^0] 
=  116°  42^  30^^ 

^  {B+A)  =  102°  47^  36^^. 

^{B-A)=    13°  54' 54^^. 

i(5  +  a)=    97°2F36.5'^ 

l{b-a)=    23°  3V  58.5^'. 

log  sin  J  {B+A)  =  9.98908 
log  CSC  J  (5-^)  =  0.61892 
log  tan  Hb-a)  =  9.63898 
log  tan  J  c  =10.24698 

^c=  60°28'43.5'^ 
c=120°57'27'^ 

log  sin  ^{b  +  a)  =  9.99641 
logcscH&-a)  =0.39873 
log  tan  ^  {B-A)  =  9.39401 

log  cot  J  C 


Exercise  XXVII.    Page  128. 

find    ^1  =  120°  47M5'^ 

B.^  =    59°  12'  15'^ 

Ci=    55°  42'    8''. 


9.78915 
58°  23^ 
C=116°47'    4''. 


Given  a  =  150°  57'    5", 

6  =  134°  15'  54", 

A  =  144°  22'  42"  ; 


Co  =    23°  57'  17.4", 


C, 


C^=   29° 


97°  42'  55", 
8'  39". 


A  >  90°,  (a  +  &)  >  180°,   a  >  6 ; 
hence  two  solutions. 

I.        sin -Sj  =  sin  .A  sin  6  CSC  a. 


log  sin  A 
log  sin  b 
colog  sin  a 
log  sin  5i 


=  9.76524 
=  9.85498 
=  0.31377 


=  9.93399 
^1=120°  47' 45". 
B^=   59°  12' 15". 


^ia-b)  =  8°  20'  35.5". 
\{a  +  b)  =  142°  36'  28.5". 
i(^-^i)=  11°  47' 28.5". 
1{A-B^)=   42°  35' 13.5". 

log  sin  i  (a  +  6)  =  9.78338 
log  esc  l{a-b)  =  0.83833 
log  tan  J  (.4-5i)=  9.31963 
log  cot  J  (7i  =  9.94134 

}Ci  =  48°51'27.7". 
Ci  =  97°  42'  55.4". 


TEACHERS     EDITION. 


143 


log  sin  i{a  +  b)  =    9.78338 

log  sin  A 

-  9.99592 

log  CSC  Ha  -  ^)  =    0.83833 

log  sin  h 

=  9.99605 

log  tan  ^  {A-B^)^    9.96338 

colog  sin  a 

=  0.00803 

log  cot  ^  (72          =  10.58509 

log  sin  B 

=  0.00000 

^0^  =  14°  34^  19.6^^ 

B  =  90°. 

C\  =  29°    8^  39^^. 

tan  c  = 

cos  A  tan  h. 

i{A  +  5i)  =  132°  35^  13.5^^. 

cot  C= 

tan  J.  cos  h. 

i  {A  +  B^)  =  101°  47^  28.5^^. 

log  cos  J. 

=    9.13499 

log  tan  h 

=  10.86812 

log  sin  H^+^i)=  9-86703 

log  tan  c 

=  10.00311 

colog  sin  ^{A-B^)=  0.68963 

c  =  45°  12^  19^^. 

log  tan  Ha- ^)  =9.16629 

log  tan  Hi           =9.72295 

Hi  =  27°  5F  4^^. 
Cj  =  55°  42^  8^^. 

log  tan  A 
log  cos  h 

log  cot  C 

=  0.86092 
=  9.12793 

=  9.98885 
C=45°44^ 

log  sin  I  {A+B.,)=  9.99074: 

colog  sin  H^-^2)=C>.16960 

log  tan  Ha -6)  =9.16629 

log  tan  H2  =  9-32663 

H2  =  11°  58^  38.7^^. 
C2  =  23°  57^  17.4^/. 

3.   Given  find 

a  =  79°  0^54.5^^  J5  =  90°, 
&  =  82°17^  4^^        c  =  45°12a9^^ 

^  =  82°   9^25.8^^;  C=45°44^ 


4.  Given  a  =  30°  52^  36.6^^,  b  = 
31°  9'  W,  A  =  87°  34^  12^'  ;  show 
that  the  triangle  is  impossible. 

From  [43] 

sin  ^  =  sin  ^  sin  b  esc  a. 


log  sin  A 
log  sin  b 
log  CSC  a 
log  sin  B 


=  9.99961 

=  9.71378 
=  0.28972 

=  0.00311 
sin  5  =1.009. 


impossible,  since  sin  5>  1. 


Exercise  XXVIII.    Page  129. 


1.   Given 
^  =  110°10^ 
5=133°18^ 
a  =  147°   5^32^^;    (7=   70°  20^  40^^. 

sin  6  =  sin  a  sin  B  esc  A. 


find 
J  =  155°  5^8^^ 
c=   33°   1^36^^ 


log  sin  a 
log  sin  B 
colog  sin  A 
log  sin  b 


=  9.73503 
=  9.86200 
=  0.02748 
=  9.62451 
h  =  155°  5^  18^^ 


144 


TRIGONOMETRY. 


^  {B  +  A)  =  121°   4:4:\ 

^{B-A)=    11°  34^ 
1(5- a)    =      3°59^53^^ 
^{b  +  a)    =  151°    5^  25^^. 

logsmH^+^)  =9.92968 

colog  sin  H^-^)  =  0.69787 

log  tan  I  (6 -a)  =8.84443 

log  tan  J  c  =  9.47198 

^  c  =  16°  30^  48^^ 
c  =  33°    1^36^^. 

colog  sin  ^{b  —  a)  =  0.15663 
log  sin  J  (5  +  «)  =  9.68433 
log  tan  J  (5-^)  =9.31104 

log  cot  J  (7  =  9.15200 

J  C  =  35°  10^  20^^. 
C=70°  20M0^^ 


2.   Given 


find 


^  =  113°39^2K^  5  =  124°  7^20^^ 
.S  =  123°40nS^^  c  =  159°53^  2^^ 
a=   65°  39^ 46^^;    (7=  159° 43^ 35^^. 

log  sin  a  =  9.95959 

log  sin  B  =  9.92024 

colog  sin  J.  =0.03812 

log  sin  b  =  9.91795 

b  =  124°  7^  20^^ 

i  (5  +  ^)  =  118°  39^  49.5^^ 
iiB-A)=      5°    0^  28.5^^. 
i  (5  -  a)    =    29°  13^  52^^. 
1  (5  +  a)    =    94°  53^  33^^. 

log  sin  1  (5+^)  =    9.94422 

colog  sin  H-S-^)  =    1.05901 

log  tan  H6  -  a)  =    9.74789 

log  tan  J  c  =10.75112 

Jc=    79°56^51^^ 
c  =  159°  53^    2^^. 


log  sin  ^-  (5  +  a)  =  9.99842 

colog  sin  i{b-a)  =  0.31128 

log  tan  i(^-^)  =8.94264 

log  cot  I  C 


=  9.25234 
1  (7=   79°5F47.7^^ 
(7  =  159°  43^  35^^ 


3.    Given  find 

J.  =  100°   2^1.3^^  5=   90°, 
B=   98°30^28^^     c=147°41M3^^ 
a=   95°20^38.7^^;C=148°  5^33^^ 


log  sin  a 
log  sin  5 
log  CSC  A 
log  sin  b 


=  9.99811 
=  9.99519 
=  0.00670 


=  0.00000 
b  =  90°. 


1{A  +  B)^  99°  16^  19.7^^. 
i{A-B)=  0°  45^51.7^^ 
Ha  -  &)    =    2°  40^  19.4^^. 

log  sin  UA+B)  =    9.99428 

colog  sin  U^-B)  =    1.87484 

log  tan  ^{a-b)  =    8.66904 

log  tan  ^c  =10.53816 

Jc=   73°  50^  51. 7^^ 
c=147°4F43^^ 

log  sin  ^  (a  +  5)  =  9.99953 

colog  sin  i{a-b)  =  1.33144 

log  tan  J  (^-5)  =8.12520 

log  cot  J  C  =9.45617 

1(7=    74°2^46.3^^ 
C=148°5^33^^ 

4.  Given  A  =  24°  33^  9^^  B  = 
38°  0^  12^^  a=  65°  20^  13^^;  show 
that  the  triangle  is  impossible. 


TEACHERS     EDITION, 


145 


cot  X  =  cos  a  tan  B. 


log  cos  a 
log  tan  B 

log  cot  X 


=  9.62042 
=  9.89286 

=  9.51328 
X  =  71°  56^  30^^ 


sin  {c  —  x)  =  cos  A  sec  B  sin  x. 


log  cos  A 
log  sec  ^ 
log  sin  X 

log  sin  (c  —  x) 
sin  (c  —  x) 


=  9.95884 
=  0.10349 
--  9.97806 


=  0.04039 
=  1.0974. 

Since  sin  {c  —  x)';>  1,  the  angle  C 
is  impossible. 

.-.  the  triangle  is  impossible. 


Exercise  XXIX.    Page  131. 

1.   Given                        find 

^  =  116°44'49'^ 

a  =  120°  55^  35^^    A  =  116°44M9^^ 

B  =    63°  15^  14^^. 

6=  59°   4^25^^    B=  63°  15^4 '^ 

C=   91°    7' 2V^, 

c  =  106°  1.0'' 22^^;     C=  91°   7^21^^ 

a  =  120°  55^  35^^ 
b=    59°    4^25'^ 
c  =  106°  10^  22^^ 

2  s  =  286°  10^  22^^ 
s  =  143°    5M1^^. 

2.   Given                        find 
a=  50°  12^  4^^    A=   59°  4^28'^ 
6  =  116°44^48^^    B=  94°23a2^^ 
c  =  129°  IF 42^^;     (7=120°   4^52^^ 

s-a^    22°    9^  36^^ 
s-b=    84°    0M6/^. 

a=    50°  12^    4^' 

s-c  =    36°  54^  49^^ 

6  =  116°  44^  48^^ 

log  sin  (s  -  a)    =    9.57657 
log  sin  (s  -  5)    =    9.99763 
log  sin  (s-c)    =    9.77859 

c  =  129°  11^  42^^ 

2s  =  296°    8^34^^ 
s  =  148°    4^  17^^ 

log  CSC  s             =    0.22141 

s_a=    97°52M3^^ 

log  tanV            =  19.57420 

s  -  5  =    31°  19^  29^^ 

log  tan  r            =    9.78710. 

s  -  c  =    18°  52^  35^^. 

tan  J  -4  =  tan  r  esc  (s  —  a). 

log  tan  J  ^         =10.21053 
log  tan  J  5        =    9.78948 

log  sin  {s-a)    =  9.99589 
log  sin  (s  -  6)    =  9.71591 

log  tan  J  (7        =  10.00851 

log  sin  (s-c)    =  9.50992 

^A=    58°  22^  24.8^^ 

log  CSC  s             =  0.27666 

^B=    31°  37.2^. 

logtan^r           =9.49838 

J  (7=    45°33M0.8^^ 

log  tan  r             =  9.74919. 

146 


TRIGONOMETEY. 


log  tan  J  ^  =9.75330 
log  tan  ^  5  =  0.03328 
log  tan  i  (7        =  0.23927 


1  A  =    29°  32^  Wr 

^B=   47°  IK  36^^ 

J  C  =    60°    2'  26^^ 

A  =    59°    4^  28^^ 

B  =    94°  23^  12'r 

C=  120°    4^  52^^, 


3,  Given  find 

a  =  131°  35^  4^^  2l  =  132°14^2F^ 

6  =  108°30a4^^  ^=110°10^40^^ 

c=   84°46^34^^;  C=   99°42^24^^ 

a  =  131°  35^  4^^ 
5  =  108°  30^  14^^ 
c  =    84°  46^  34^' 


2s  =  324°51^52^^ 

s  =  162°  25'  56'^ 

s-a=^    30°  50'  52''. 

s-b=    53°  55'  42". 

s-c  =    77°  39'  22". 


log  sin  (s  —  a) 
log  sin  (s  —  b) 
log  sin  (s  —  c) 

log  CSC  s 

log  tan^  r 
log  tan  r 


9.70991 
9.90756 
9.98984 
0.52023 

10.12754 
10.06377. 


log  tan  J  A  =  0.35386 
log  tan  J  5  =  0.15621 
log  tan  ^  (7        =  0.07393 

iA=    66°    7^  10.6". 
I  5  =   55°    5'  20". 
I  C  =   49°  51/  12". 


A  =  132°  14'  21". 
^=110°  10' 40". 
C=    99°  42' 24". 

4.   Given  find 

a  =  20°16'38",  A=   20°   9' 54", 

6  =  56°  19'  40",  B  =   55°  52'  31", 

c  =  66°  20'  44"  ;  C  =  114°  20'  17". 

a  =  20°  16'  38" 
b  =  56°  19'  40" 
c  =    66°  20'  44" 


2s  =  142°  57'    2" 

s  =    71°  28'  31". 

s-a=    51°  11' 53". 

s-b=    15°    8' 51". 

s-c=     5°    7' 47". 


log  sin  (s  —  a) 
log  sin  (s  —  b) 
log  sin  (s  —  c) 

log  CSC  s 

log  tan^  r 
log  tan  r 


9.89172 
9.41715 
8.95139 
0.02311 

8.28337 
9.14168. 


log  tan  M  =  9.24996 
log  tan  J  5  =  9.72453 
log  tan  i  (7        =  10.19029 

J  J.  =  10°  4'  56.8". 
J  5  =  27°  56'  15.5". 
f  (7=  57°  10' 8.6". 

A  =  20°  9'  54". 

.5  =  55°  52' 31". 

a=  114°  20'  17". 


TEACHERS     EDITION. 


147 


•    Exercise  XXX.     Page  132. 


1.   Given 

find 

^  =  130°, 

a  =  139°2F22^^ 

£=110°. 

6  =  126°  57^  52^^ 

C=    80°; 

c=  56°51M8^/. 

A  = 

130° 

B  = 

110° 

c  = 

80° 

2^  = 

320° 

s  = 

160°. 

S- 

A  = 

30°. 

S- 

B  = 

50°. 

S- 

-C  = 

80°. 

log  cos  S 

=    9.97299 

log  sec  (^S^- 

-A) 

=    0.06247 

log  sec  {8- 

-B) 

=   0.19193 

log  sec  {S- 

-c) 

=    0.76033 

log  tan^  E 

=  10.98772 

log  tan  B 

=  10.49386. 

log  tan  ^  a 

=  10.43139 

log  tan  ^  b 

=  10.30193 

log  tan  ^  c 

la 

=    9.73353 

=    69°40M1^^ 

JS 

=    63°  28-' 56'^. 

i^ 

=    28°25^54'^ 

a 

=  139°  2V  22''. 

h 

=  126°  57^  52'^. 

c 

-   56°5F48''. 

2.   Given 

find 

A  =  59°5Wl(y^, 

«i=128°42^29'^ 

i?  =  85«36^50^^ 

h=   64°  2^47^^ 

c=^d°mn(y^; 

c=  128°  42' 29'^. 

A  =  59°  55^  10^^ 
B  =  85°  36^  50^^ 
C  =    59°  55^  10^^ 

2^=205°  27^  10^^ 

S  ==  102°  43'  35'\ 

iS-A=   42°  48^  25^^. 

S-B=    17°    6M5^^. 

S-C^   42°48^25^^ 

log  cos  S  =  9.34301 

log  sec  {S-A)  =0.13451 
log  sec  (S-B)  =0.01967 
log  sec  {S-C)  =0.13451 
log  tan^i?  =  9.63170 

log  tan  E  =  9.81585. 


log  tan  I  a 
log  tan  I  b 
log  tan  I  c 


=  9.68134 

=  9.79618 
=  9.68134 


^  a  =  25°  38^  45.5^^. 
^6=32°  y23.6^^. 
^  c  =^  25°  38^  45.5^^ 

a=-51°17^3F^. 

J  =  64°  2M7^^. 

c  =  51°17^3F^. 

3.   Given  find 

A  =  102°  W 12'',  a  =  104°  25^  9", 

B=  54°  32' 24:",  b=  53°  49^  25^^, 

C=  89°   5^46^^;  c=   91°  44' 24". 

A  =  102°  14^  12^' 
B  =  54°  32^  24^^ 
C=    89°    5M6^^ 


2  ^=245°  52^  22^/ 
^S'=122°  50^  11^^, 


148 


TRIGONOMETRY. 


8-A- 

=  20°  41^  59^^. 

A  = 

4°  23^  35^^ 

S-B- 

=  68°  23^  47^^ 

B  = 

8°  28^  20^^ 

S-C  = 

=  33°  50^  25^^ 

C  = 

172°  1*7^  56^^ 

log  cos  S 
log  sec  {S 
log  sec  {S 
log  sec  {S 
log  tan^iJ 

=  9.73536 
~A)=  0.02898 
-B)  =  0.43394  • 
-  (7)  =  0.08061 

=  0.27889 

2/S'  = 

S= 

S-A  = 

S-B  = 

S-C==- 

185°    9^51^^ 

92°  34^  55.5^^. 

88°  11^  20.5^^ 

84°    6^  35.5^^. 

-(79°  43^    0.5^0- 

log  tan  a 

=  0.13945. 

log  cos  S 

=    8.65368 

log  tan  ^  ( 
log  tan  ^  < 

z        =0.11047 
b        =  9.70551 

log  sec  {S  — 
log  sec  {S  — 
log  sec  (-S*  — 

A)  =    1.50029 

B)  =    0.98876 

C)  =    0.74833 

log  tan  ^ 

c         =  0.05885 

log  tan^i^ 

=  11.89106 

la=    52°  12^  34.6^^. 

log  tan  E 

=  10.94553. 

ih=    26°  54^  42.5^^. 

ic=   48°  52^  12^^ 
a  =  104°  25^    9^^ 
6  =    53°  49^  25^^. 
c=    97°44^24^^ 

log  tan  J  a 
log  tan  J  b 
log  tan  ^  c 

=    9.44524 
=    9.95677 
=  10.19720 

la=    15°  34^  35.5^^ 
J6=    42°    9^  11.5^^ 

4.  Given 

find 

Jc=    57°34^58^^ 

A=     4°  23^ 

35^^    a=  31°  9^1^^ 

a=    31°    9MF^. 

^=     8°  28^ 

20^^,     b=   84°18^23^^ 

b  =    84°  18^  23^^ 

C  =  172°17^ 

56^^;     c  =  115°  9^56^^ 

c  =  115°    9^56^^ 

Exercise  XX> 

[I.     Page  135. 

1.  Given 

find 

log  26159 

=  4.41762 

A  =  84°  20^ 

19^^      ^=26159^^ 

colog  648000 

=  4.18842-10 

B  =  27°  22^ 
C  =  75°  33^; 

40^^      i^=  0.12685  i22. 
=  ^+5+C-180°. 

log  3.14159 
logi^ 

=  0.49715 

=  9.10319-10 

A  = 

=    84°  20^  19^^ 

i^=  0.12685  i^^ 

B  = 

=    27°  22^  40^^ 

G  = 

-    75°  33^ 

187°  15^  59'^ 

2.   Given 

find 

180° 

a=  69°  15^  6 
6  =  120°  42^  47 

'',    ^=216°40a8^^. 

E  = 

-     7°  15^  59^^ 

=  26159^^ 

c=  159°  18^  33 

TEACHERS     EDITION. 


149 


a  =  69°  15^ 
6  =  120°  42^ 
c  =  159°  18^ 


6'^ 

47// 


2  s  =  349°  16^ 

s  =  174°  38^ 

s-a  =  105°  23^ 

s-b=    53°  55^ 

s-c=    15°  19^ 


^s 


=  87°  19^ 

-^  (s  -  a)  =  52°  41^ 

Hs-6)=  26°  57^ 

1  (s  -  c)  =  7°  39^ 


26^/ 
13^^. 

26^^ 
40^^. 

6.5^^ 
33.5^^ 
43^^ 
50^^. 


log  tan  ^s  =11.32942 

log  tan  |(s- a)  =10.11804 
log  tan  J(s-6)=  9.70645 
logtan  ^(s-c)=  9.12893 
logtan^J.E'  =10.28284 
log  tan  i~E       =  10.14142. 

IE=    54°  10^    4.6^^ 
-£;=216°  40^  18^^. 

3.   Given  find 

a=   33°   1^45^^   ^=133°  48^  53^^. 
6  =  155°  5a8^^ 
(7  =  110°  10^; 

tan  m  =  tan  a  cos  C.  (^  54) 

cose   =  cos  a  seem  cos  (6— m).    (^54) 

log  tan  a  =  9.81300 

log  cos  c  =  9.53751 

log  tan  m  =  9.35051 

m=    167°  22^ 

6-m  =  -(12°16^42^0- 

log  cos  a  =  9.92345 

log  sec  m  =  0.01064 

log  cos  (6  -  m)  =  9.98995 

log  cos  c  =  9.92404 

c  =  147°  5^  30^^ 


a=  33°  1M5^^ 
h  =  155°  5^  18^^ 
c  =  147°    5^  30^^ 


2s  = 

335°  12^  33^/ 

s  = 

167°  36^  16.5^^. 

s  —  a  = 

134°  34^  31.5^^. 

s-6  = 

12°  30^  58.5^^. 

s  —  c  = 

20°  30^  46.5^^. 

^s  = 

83°  48^    8.25^^ 

Hs  -  «)  = 

67°  17^  15.75^^ 

Hs-6)  = 

6°  15^  29.25^^ 

Hs-c)  = 

10°  15^  23.25^^ 

log  tan  J  s  =  0.96419 

log  tan  J  (s- a)  =0.37824 
log  tan  ^(s- 6)  =9.04005 
log  tan  i{s  —  c)  =  9.25755 


log 

tan^ 

IE 

9.64003 

log 

tan 

iE 

9.82002. 

i 

E  = 

33°  27^ 

13^^/ 

E  = 

133°  48^ 

53^^. 

4.   Find  the  spherical  excess  of  a 
triangle  on  the  earth's  surface  (re- 
garded as  spherical),  if  each  side  of 
the  triangle  is  equal  to  J.°. 
Given  a,  6,  and  c  each  =  1°;  then 
2s  =  3°.  ^s  =  45^ 

1°  30^      ^(^s-a)  =  15^ 


Hs-6)  = 
Hs-c)  = 

=    8.11696 


s 
s-a=  30^ 
s-h=  30^ 
8-c=  30^. 
log  tan  J  s 
log  tan  J  (s- a)  =  7.63982 
logtan  J(s-6)=  7.63982 
log  tan  J  (s-c)  =    7.63982 

logtanH^       =11.03642 
log  tan  i  ^       =    5.51821. 
i.£'=6.814^^ 
j5'=27.25^^ 


15^. 
15^. 


150 


TRIGONOMETRY. 


Exercise  XXXII.     Page  148. 


1.  Find  the  dihedi-al  angle  made 
by  the  lateral  faces  of  a  regular  ten- 
sided  pyramid ;  given  the  angle  A 
=  18°,  made  at  the  vertex  by  two 
adjacent  lateral  edges. 


About  0  the  vertex  of  the  pyra- 
mid, describe  a  sphere.  It  will 
intersect  the  lateral  surface,  forming 
a  regular  spherical  decagon,  of  which 
the  sides  =  18°,  being  measured  by 
the  plane  angles  at  the  centre. 

Pass  a  plane  through  0  and  A 
and  C,  forming  an  isosceles  spheri- 
cal triangle  ABC. 

Then  (by  Prob.  3,  Ex.  XXIII.) 
given  side  a  =  18°,  n  =  10,  to  find 
angle  A  of  the  polygon. 

■    1  A            I  A        180° 
•sm  ^  A  =  sec  ^  A  cos -• 


log  cos  18^ 
colog  cos  9° 


=  9.97821 
=  0.00538 


log  sin  M      =  9.98359 
74°  21^ 
A  =  148°  42^ 


iA 


2.  Through  the  foot  of  a  rod 
which  makes  the  angle  A  with  a 
plane,  a  straight  line  is  drawn  in 
the  plane.  This  line  makes  the 
angle  B  with  the  projection  of  the 


rod  upon  the  plane.     What  angle 
does  this  line  make  with  the  rod  ? 


Let  CO  be  a  straight  line,  making 
the  angle  A  with  the  plane  OH; 
IF  Sb  straight  line  passing  through 
the  foot  of  CO,  making  the  angle  B 
with  the  projection  DO  of  CO  upon 
the  plane  GH. 

It  is  required  to  find  the  angle 
COI==  X. 

AVith  a  radius  equal  to  unity, 
from  0  as  a  centre,  construct  the 
spherical  triangle  DCI. 

Then  i  =  A, 

c=B, 

d=COI=x 

CEK=-Tt.  angle. 

Since  OE  is  the  projection  of  CO 
on  the  plane  OH;  CE  drawn  from 
C  to  E,  is  perpendicular  to  OE. 

.'.  CDI  =rt.  triangle. 

By  Formula  [37], 

cos  d  =  cos  i  cos  c. 

.•.  cos  X  =  cos  A  cos  B. 

3,  Find  the  volume  V  of  an  ob- 
lique parallelopipedon ;  given  the 
three  unequal  edges  a,  b,  c,  and  the 


TEACHERS     EDITION, 


151 


three  angles  I,  m,  n,  which  the  edges 
make  with  one  another. 


/\ 

r 

\ 

">-<^ 

\¥^, 

\ 

a 

N^ 

__, 

-^■-"^A 

C 

B 


Let  ^5  be  a  parallelopipedon, 
and  I,  m,  and  n,  the  angles  which 
the  unequal  edges  a,  b,  and  c  make. 

We  are  to  find  V,  the  volume. 

Let  w  =  the  inclination  of  the 
edge  to  the  plane  of  a  and  b. 

V=  area  base  X  altitude. 
Area  base  =^  bz,  when  z  =  a  sin  I. 
Altitude     =  X,   when  a?  =  c  sin  w. 
.'.  V=  abc  sin  I  sin  w. 

Suppose  a  sphere  to  be  constructed 
having  for  its  centre  the  vertex  of 
the  trihedral  angle  whose  edges  are 
a,  b,  and  c.  The  spherical  triangle 
whose  vertices  are  the  points  where 
a,  b,  and  c  meet  the  surface  has  for 
its  sides  I,  m,  n ;  and  vj  =  perpen- 
dicular arc  from  side  I  to  the  oppo- 
site vertex. 

Let  L,  M,  iV  denote  the  angles  of 
the  triangle. 

Then  by  [38]  and  [47], 

sin  w  =  sin  m  sin  N 

=  2  sin m sin  J  iVcos  |  N, 
Or  if   s  =  l{l-\-m^-n), 
sin  w  = 

- —  Vsin  s  sin(s— Z)  sin(s— m)  sin(s— n) 
sin? 

.-.  V= 

2a6c  Vsinssin(s— Z)  sin(s— m)  sin(s— ri) 


4.  The  continent  of  Asia  has 
nearly  the  shape  of  an  equilateral 
triangle,  the  vertices  being  the 
East  Cape,  Cape  Romania,  and  the 
Promontory  of  Baba.  Assuming 
each  side  of  this  triangle  to  be  4c5UO 
geographical  miles,  and  the  earth's 
radius  to  be  3440  geographical 
miles,  find  the  area  of  the  triangle : 

(i.)  regarded  as  a  plane  triangle  ; 

(ii.)  regarded  as  a  spherical  triangle. 

Area        =  J  (base  x  altitude). 

Altitude  =  V48002  -  24002. 

=  V 17280000. 


log  \/l7280000=  3.61877 
log  2400  =3.38021 

log  area  =6.99898 

Area  =  9976500. 


(ii.) 


180° 


a,  b,  and  c 


4800° 
60 


2  s  =  240° 

Hs- 

-a)=    20° 

i(s- 

-6)=    20° 

Hs- 

-c)=    20° 

log  tan  J  s  =  10.23856 

log  tan  1  (s  -  a)  =  9.56107 
log  tan  l{s-b)  =  9.56107 
log  tan  l{s-c)  =  9.56107 
logtan^i^         =    8.92177 

1^=16°    7^    ^.V\ 
J5'=64°  28^32.5^^ 
=  232112.5^^ 


152 


TRIGONOMETRY. 


logi: 
log- 


'6-18000 
logi?2 

logi^ 


=  5.36570 

=  4.68557 

=  7.07312 

=  7.12439 

i^=  13316560. 


5.  A  thip  sails  from  a  harbor  in 
latitude  I,  and  keeps  on  the  arc  of  a 
great  circle.  Her  course  (or  angle 
between  the  direction  in  which  she 
sails  and  the  meridian)  at  starting 
is  a.  Find  where  she  will  cross  the 
equator,  her  course  at  the  equator, 
and  the  distance  she  has  sailed. 


Let  NESW  be  the  earth,  WCE 
the  equator,  iVand  Sthe  nortli  and 
south  poles.  Let  A  be  the  point 
from  which  the  ship  starts,  AFD 
the  parallel  of  latitude  the  ship 
starts  from,  and  AB  the  great  circle 
of  its  course. 


Then 
BAE=a 

=  course  of  ship. 

log  sin  \  A 
\A 

=  10.00000  - 
=  90°. 

-10 

AE=l 

=  latitude  of  its  starting- 

A 

=  180°. 

place. 

Arc    a 

=  180°. 

BE  =  TO  =  place   of  crossing  the 
equator. 

B  =  course  at  equator. 

AB  =  c?  =  distance  sailed. 

By  Napier's  rule, 

sin  I   —  tan  m  cot  a  ; 

whence     tan  m  =  sin  I   tan  a. 

cos  B  =  cos  I  sin  a. 

cot  d  =  cot  I  cos  a. 

6.  Two  places  have  the  same 
latitude  I,  and  their  distance  apart, 
measured  on  an  arc  of  a  great  cir- 
cle, is  d.  How  much  greater  is  the 
arc  of  the  parallel  of  latitude  be- 
tween the  places  than  the  arc  of  the 
great  circle  ?  Compute  the  results 
for  I  =  45°,  d  =  90°. 


In    isosceles    spherical     triangle 
ABO 

sin  J  J.  =  sin  *  dcsc{dO°-l) 
=  sin  J  c?sec  I. 
Let  ?=45°,   d=90°. 

log  sin  J  d    =    9.84949 
log  sec  I       =    0.15051 


TEACHERS     EDITION. 


153 


Arc  K  =  a  X  cos  Z 

=  iaV2  =  90°\/2. 
90°  V2  -  90°  =  90°(  V2  - 1). 

7.  The  shortest  distance  d  be- 
tween two  places  and  their  latitudes 
I  and  V  are  known.  Find  the  dif- 
ference between  their  longitudes. 


Let  C  represent  the  north  pole, 
A  the  position  of  one  city,  B  the 
position  of  the  other.  Then  AB  =  d, 
and  if  m  represent  the  longitude  of 
one  city,  mV  that  of  the  other,  I  the 
latitude  of  one  city,  and  V  that  of 
the  other,  angle  Cwill  be  equal  to 
{m-m^),  and  5(7  will  equal  90°- Z, 
and  ^Cwill  be  equal  to  90° -^^ 

Therefore  we  have  an  oblique 
spherical  triangle  with  three  sides 
given  to  find  the  angle  C. 

Now  from  Formula  [44], 
cos  c  =  cos  a  cos  h  +  sin  a  sin  h  cos  C; 
then  by  substituting, 

cos  d  =  cos  (90°-  I)  cos  (90°  -  V) 
+  sin  (90°  -  I) 
X  sin  (90°—  V)  cos  (m  —  m^) ; 
or   cos  d  =  sin  I  sin  V 

+  cos  I  cos  V  cos  (m  —  m^). 

.'.  cos  (w  —  m^)  =  (cos  (i  —  sin  Z  sin  V) 

X  sec  I  sec  V. 


8.  Given  the  latitudes  and  longi- 
tudes of  three  places  on  the  earth's 
surface,  and  also  the  radius  of  the 
earth  ;  show  how  to  find  the  area 
of  the  spherical  triangle  formed  by 
arcs  of  great  circles  passing  through 
the  places. 


Let  A,  5,  and  G  represent  the 
positions  of  three  places  on  the 
earth's  surface. 

I  62  shows  how  to  find  the  dis- 
tance between  two  places  when  the 
latitudes  and  difference  in  longitude 
are  given. 

In  this  case  we  have  the  latitudes 
given,  and  also  the  longitudes ;  so 
that  we  can  find  the  difference  in 
longitude. 

Let    GH  =  equator  ; 

then,  from  |  54,  in  triangle  ABC, 

tan  m  =  cot  a  cos  6, 

and  from  §  62, 

cos  BC=  sin  a  sec  m  sin  (5+m), 

and  the  same  with  the  distance  be- 
tween the  other  places. 

Therefore,  we  have  given  the 
three  sides  of  a  spherical  triangle, 
to  find  the  area. 


154 


TRIGONOMETRY. 


By  [51], 

tan^ \  E=  tan  I  s  tan i (s  —  a) 

X  tan  i{s  —  h)  tan  J  (s  —  c). 

Then,  since  we  have  the  radius  of 
the  sphere  given  (=  R)  and  the 
spherical  excess  =  E,  from  Formula 

F=^TrR\ 


180= 

9.  The  distance  between  Paris 
and  Berlin  (that  is,  the  arc  of  a 
great  circle  between  these  places)  is 
eq'ial  to  472  geographical  miles. 
The  latitude  of  Pans  is  48°  50^  13'^; 
that  of  Berlin,  52°  30^  16^^.  When 
it  is  noon  at  Paris  what  time  is  it 
at  Berlin  ? 


Let  AO  represent  the  latitude 
of  Paris,  and  BK  the  latitude  of 
Berlin.  Then  C  represents  the  dif- 
ference in  longitude. 

6M  =  6  =  41°  9M7^^ 
CB  =  a=2>T  29M4^^ 
AB==c=    7°  52^  (472  H- 60) 

2s=  86°  31^31'/ 

s  =  43°  15^  45.5^^ 

s-a=    5°  46^    1.5^^ 

s-b=   2°    5^58.5>'>'. 

s  -  c  =  35°  23^  45.5^^ 

tan'^  J  C=  cscs 

sin  (s  —  a)  sin  (s  —  b)  esc  (s  —  c). 


log  CSC  s  =  0.16409 

log  sin  (s- a)  =  9.00210 

log  sin  (s- 5)  =  8.56391 

log  CSC  (s-c)  =  0.23716 

log  tannic       =17.96726 
tan  J  (7       =    8.98363 
1(7=5°  30^  2^^ 
(7-11°    0^4^^ 

-  660. 

Difference  of  time 

=  Y^5(660)  minutes 
=  41  min. 

Time  at  Berlin,  12  h.  44  mm. 

10.  The  altitude  of  the  pole  be- 
ing 45°,  I  see  a  star  on  the  horizon 
and  observe  its  azimuth  to  be  45°  ; 
find  its  polar  distance. 


iVP=45°.  FM^p. 

.:PZ  =  4:5°  =  1.         PZM^a. 

We  have  given  two  parts  of  the 
triangle, 

a  =  45°. 

(90°  -  0  =  45°. 

cosp  =  cos  a  cos  I. 

cos  a  =  \/|. 

cos  I  =  V^. 
.'.  cosp  =  J. 
p  =  60°. 


TEACHEES     EDITION. 


155 


11.  Given  the  latitude  I  of  the 
observer,  and  the  declination  d  of 
the  sun  ;  find  the  local  time  (appar- 
ent solar  time)  of  sunrise  and  sunset, 
and  also  the  azimuth  of  the  sun  at 
these  times  (refraction  being  neg- 
lected). When  and  where  does  the 
sun  rise  on  the  longest  day  of  the 
year  (at  which  time  d=  -\-  23°  27^) 
in  Boston  {I  =  42°  2V),  and  what  is 
the  length  of  the  day  from  sunrise 
to  sunset?  Also,  find  when  and 
where  the  sun  rises  in  Boston  on 
the  shortest  day  of  the  year  (when 
(^=- 23°  270,  and  the  length  of 
this  day. 

To  find  the  hour  angle  t  when  the 
sun  is  on  the  horizon. 


PM:=  90°  -  d. 
^Q  =  90°. 
.-.  PQ  =  90°  -  Z  -h  90° 
=  180°  - 1. 

Then  in  triangle  PMQ,  by  [39], 
cos  QPM==  tan  PQ  cot  Pif, 
or       cos  t  =  tan(180°-r)cot(90°-tf), 
cos  t  =  ~  tan  I  tan  d. 


Also  to  find  azimuth  a. 

In  triangle  PMQ,  MQ  is  meas- 
ured by  angle  QZM=  180°  -  a. 

Then,  by  [37], 
cos  PM=  cos  PQ  cos  MQ. 
cos  (90°  -  d) 

=  cos  (180° -  Z)  cos  (180°  -  a). 
sind  =  —  cos  I  (—  cos  a), 
or       cos  a  =  sin  c?  sec  I, 

Now  cos  t  =  —  tan  d  tan  I. 

Time  of  sunrise  =  12 o'clk  a.m. 

15 


Time  of  sunset 


15 


o'clk  P.M. 


log  tan  d  =  9.63726 
log  tan  I  =  9.95977 
log  cos  t  =  9.59703 

t  =  66°  42^  26^^ 

12  -  —  =  7  h.  33  min.  10  sec. 
15 


15 


=  4  h.  26  min.  50  sec. 


.*.  shortest  day 

=  2  X  4  h.  26  min.  50  sec. 
=  8  h.  53  min.  40  sec. 
Again, 

cos  a  =  sin  c?  sec  I. 
log  sin  d  =  9.59983 
log  sec  I    =  0.13133 
log  cos  a  =  9.73116 

a  =   57°  25^  15^^ 
.-.  a^=122°  34^45^^ 
Longest  day  =  12  hrs. 

-h[(7h.  33min.  lOsec. 

-4h.  26  min.  50  sec.)] 
=  15  h.  6  min.  20  sec. 


156 


TRIGONOMETKY. 


12.  AVlien  is  the  solution  of  the 
problem  in  Example  11  impossible, 
and  for  what  places  is  the  solution 
impossible  ? 

d  has  for  its  maximum  value 
23°  27^. 

Suppose      I  =  66°  Sy. 
Then  tan  (180°  -  Z)  =  -  tan  ^ 
=  —  cot  d. 
Formula  cos  t  =  —  tan  I  tan  d 
becomes   cos  t^  —  cot  d  tan  d 
=  -1. 
.-.  i=180°; 

that  is,  the  sun  just  appears  in  the 
south  on  shortest  day. 

For  places  within  the  Arctic  cir- 
cle, I  >  66°  33^,  and  —tan  I  numeri- 
cally greater  than  —  cot  d. 

Hence    —  tan  I  tan  c?  >  —  1 

(numerically),  or 

cos  t  =  ±  1+, 

which  is  not  possible.     Hence   the 
sun  may  fail  to  rise  during  24  hours. 

13.  Given  the  latitude  of  a  place 
and  the  sun's  declination ;  find  his 
altitude  and  azimuth  at  6  o'clock 
A.M.  (neglecting  refraction).  Com- 
pute the  results  for  the  longest  day 
of  the  year  at  Munich  {I  =  48°  9^. 

FZM=  a. 
FZ=dO°-l 
PJf=90°  -d=p. 

ZPM=  t. 
ZM=  90°  -  h. 
I  =  48°  9^ 

Sun's  declination  on  longest  day, 
23°  27^. 


By  Napier's  Rules, 

sin  h  =&hxl  sin  d. 
log  sin  I  =  9.87209 
log  sin  d  =  9.59983 
log  sin  h  =  9.47192 
Altitude  =  A  =  17°  14^  35^^ 

By  Napier's  Rules, 

cot  a  =  cos  7  tan  d. 

log  cos  I  =  9.82424 

log  tan  d  =  9.63726 

log  cot  a  =  9.46150 

Azimuth  =  a  =  73°  5V  W\ 

14.  How  does  the  altitude  of  the 
sun  at  6  a.m.  on  a  given  day  change 
as  we  go  from  the  equator  to  the 
pole  ?  At  what  time  of  the  year  is 
it  a  maximum  at  a  given  'place  ? 
(Given  sin  A  =  sin  Z  sin  d.) 

The  farther  the  place  from  the 
equator,  the  greater  the  sun's  alti- 
tude at  6  A.M.  in  summer.  At  the 
equator  it  is  0°.  At  the  north  pole 
it  is  equal  to  the  sun's  declination. 
At  a  given  place,  the  sun's  altitude 


TEACHERS     EDITION". 


157 


at  6  A.M.  is  a  maximum  on  the 
longest  day  of  the  year,  and  then 
sin  h  =  sinl sin  e  (where  e  =  23° 27''). 

15.  Given  the  latitude  of  a  place 
north  of  the  equator,  and  the  dec- 
lination of  the  sun  ;  find  the  time  of 
day  when  the  sun  bears  due  east 
and  due  west.  Compute  the  results 
for  the  longest  day  at  St.  Peters- 
burg {I  =  59°  560- 


Let  NESWhQ  the  horizon,  ^  the 
zenith,  NZS  the  meridian,  WZE 
the  prime  vertical,  WAE  the  equi- 
noctial, P  the  elevated  pole,  M  the 
position  of  the  sun  when  due  east, 
MB  its  declination,  and  ZPM  its 
hour  angle. 

Then  we  have  the  right  spherical 
triangle  PZM,  with  PM  and  PZ 
known,  to  find  ZPM. 

Given  I  and  c?,  to  find  U 
PM=  90°  -  d. 
PZ=dO°-l 

From  ^  48,  Case  II., 
cos 5=  tana  cote. 

Substitute  in  this  equation, 
cos  t  =  tan  PZ  cot  PM, 
or     cos  t  =  tan  (90°- 1)  cot  (90°-  d). 
.'.  cos  t  =  cot  I  tan  d. 


And  from  ^  65,  the  times  of  bear- 
ing due  east  and  west  are 

12 A.M.   and  —  p.m., 

15  15 

respectively. 

Since  the  day  given  is  the  longest 
day  of  the  year,  the  declination  of 
the  sun  =  23°  27^ 

.*.  we  have  given  d=  23°  27''  and 
I  =  59°  56^  to  find  t. 

Now  cos  t  =  cot  I  tan  d. 

log  cot  I  =  9.76261 
log  tan  cZ=  9.63726 

log  cos  t  =  9.39987 

t  =  75°  27^  24^^ 

.*.  12 =  6  hrs.  58  min.  A.M., 

15 

and  —  =  5  hrs.    2  min.  p.m. 

15 

16.  Apply  the  general  result  in 
Example  15  (cos  t  =  cot  I  tan  d)  to 
the  case  when  the  days  and  nights 
are  equal  in  length  (that  is,  when 
d  =  0°).  Why  can  the  sun  in  sum- 
mer never  be  due  east  before  6  a.m., 
or  due  west  after  6  p.m.  ?  How 
does  the  time  of  bearing  due  east 
and  due  west  change  with  the  decli- 
nation of  the  sun  ?  Apply  the  gen- 
eral result  to  the  cases  where  I  <Cd 
and  1=  d.  What  does  it  become  at 
the  north  pole  ? 

When  the  days  and  nights  are 
equal,  c?=0°,  cosi=0°,  and  ^  =  90°; 
that  is,  the  sun  is  due  east  at  6  a.m. 
and  due  west  at  G  p.m.  Since  I  and 
d  must  both  be  less  than  90°,  cos  t 
cannot  be  negative ;  therefore  t  can- 
not be  greater   than   90°.      As   d 


158 


TRIGONOMETRY. 


increases,  t  decreases ;  that  is,  the 
times  in  question  both  approach 
noon. 

lfl<d,  then  cos  i >  1 ;  therefore 
this  case  is  impossible.  If  1  =  d, 
then  cost=l,  and  t  =  0°;  that  is, 
the  times  both  coincide  with  noon. 

The  explanation  of  this  result  is, 
that  the  sun  at  noon  is  in  the 
zenith ;  hence,  on  the  prime  vertical, 
at  the  pole,  Z  =  90°,  cosi=  0°, 
i  =  90°;  therefore  the  sun  in  sum- 
mer always  bears  due  east  at  6  a.m. 
and  due  west  at  6  p.m. 

17.  Given  the  sun's  declination 
and  his  altitude  when  he  bears  due 
east;  find  the  latitude  of  the  ob- 
server. 

N 


In  the  figure  let  Z  be  the  zenith 
of  observer,  F  the  elevated  pole,  M 
the  position  of  the  sun. 

Then  ZM=  90°  -  h. 
FM=  90°  -  d. 
FZ=dO°-l. 

Since  the  sun  M  bears  due  east, 
MZF  is  a  right  angle. 

/.  by  Napier's  Kules, 

cos  FM=  cos  FZ  cos  MZ. 


.'.  sin  d^sinl  sin  h. 
sin  I  =  sin  dcsch. 

18.  At  a  point  0  in  a  horizontal 
plane  MI^  a  staff  OA  is  fixed,  so 
that  its  angle  of  inclination  AOB 
with  the  plane  is  equal  to  the  lati- 
tude of  the  place,  51°  30^  N.,  and 
the  direction  OB  is  due  north. 
What  angle  will  OB  make  with  the 
shadow  of  OA  on  the  plane,  at  1 
P.M. 


Given  direction  of  OB  due  north, 
AOB  =  51°  30^=  I,  and  plane  MN 
horizontal ;  to  find  BOC. 

Produce  OA  ;  it  will  pass  through 
the  pole.  The  sun  being  on  the 
equinoctial,  FOS  =  90°,  and  the 
shadow  00  will  lie  in  the  plane  of 
this  angle.  Draw  OZ  JL  to  plane 
MJSF;  it  will  lie  in  the  plane  of  OB 
and  OA. 

SFZ  =  hour  angle  of  sun  at  1  p.m. 
=  15°. 

SFZ  =  CAB,  being  vertical  angles. 
.■.CAB=\b°. 

ABO=  90°,  since  OB  is  the  pro- 
jection of  OA  on  plane  MN. 

Arc  AB  =  51°  30^,  being  the  meas- 
ure of  plane  angle  AOB. 

Then  in  right  spherical  triangle 
ABC,  by  [41], 

tan  BC^  tan  BAGs'm  AB. 


TEACHERS     EDITION. 


159 


log  tan  15°  =9.42805 
log  sin  51°  30^  =  9.89354 
log  tan  BQ  =  9.32159 
Arc  BO  =  11°  50^  35^^ 

Arc  -SC measures  plane  angle  BOC. 
.:  BOC=  11°  50^  35^^ 

19.  What  is  the  direction  of  a 
wall  in  latitude  52°  30^  N.  which 
casts  no  shadow  at  6  a.m.  on  the 
longest  day  of  the  year. 


The  wall  must  lie  in  the  line  pass- 
ing through  the  sun,  in  order  that  it 
may  cast  no  shadow. 

In  the  figure, 

PZ=  90° -I 
PM=  90°  -  d. 
■     ZPD  =  6  X  15°=  90°. 
To  find  irZP=a;. 
By  [41], 
sin  (90°-  V)  =  tan  {90° -d)  cot  MZP. 

cos  I  =  cot  d  cot  X. 
Or,       cot  re  =  cos  Z  tan  e. 

log  cos  I   =  9.78445 
log  tan  e  =  9.63726 

log  cot  a;  =9.42171 

X  =  75°  12^  38^^. 


20.  At  a  certain  place  the  sun  is 
observed  to  rise  exactly  in  the  north- 
east point  on  the  longest  day  of  the 
year  ;  find  the  latitude  of  the  place. 

When  the  sun  rises  in  the  north- 
east on  the  longest  day  of  the  year, 
a  =  45°,  c?=23°  27^ 

To  find  I.     In  the  formula 
cos  a  =  sin  d  sec  I. 

log  sec  I  =  log  cos  a  +  log  esc  d. 

log  cos  45°  =  9.84949 
log  CSC  23°  27^  =  0.40017 
log  sec  I  =  0.24966 

I  =  55°  45^  6^^. 

21.  Find  the  latitude  of  the  place 
at  which  the  sun  sets  at  10  o'clock 
on  the  longest  day. 

Since  the  sun  sets  at  10  o'clock, 
the  hour  angle  of  the  sun  is  equal 
to  15°  X  10  =  150°. 

The  declination  of  the  sun  is  equal 
to  23°  27^. 

.'.  we  have  in  the  triangle  two 
parts  given ;  viz.,  the  angle  ZPM, 
and  ifP=  90° -d 

In  the  formula, 

cot  I  =  cos  t  cot  d. 
t  =  150°. 
d=   23°  27^ 

log  cos  t  =  9.93753 
log  cot  £?  =  0.36274 

log  cot  I  =  0.30027 

I  =  63°  23^  41^^ 

22.  To  what  does  the  general 
formula  for  the  hour  angle,  in  ^  67, 
become  when  (i.)  h  =  0°,  (ii.)  I  =  0° 
and  c?  =  0°,  (iii.)  Zor  c?  =  90°. 


160 


TEIGONOMETRY. 


In  the  general  formula,  §  67,  let 
/i  =  0.     Then 

sin  I  i  =  ±  [cos  J(Z  ^- p) 

X  sin  J  (Z  +p)  sec  I  cscp]^. 


(11.) 


sin  i  t 


^  h  —  cos  t 


cos 


j(;+rt  =  .^I±£2^(U£). 


sin  I 


ii+p)^^^j'-^°f^pl 


sec  Z  cscp     = 


cos  I  sinp 


Substitute  these  values  in  the  first 
equation, 

1  —  cos  ^  ^     /l  +cos(?  +  p) 
2        ~^  2 

^  Jl_-cos_(Z_+^) 


X 


cos  I  sinp 


1  —  cos  i  =  Vl  —  cos^  {I  +p) 

COS  I  smp 
1  —  cos  i  =  sin  (I  +  p) 

COS  Isinp 
=  (sin  I  cosp  +  COS  I  sinp) 

x^L_ 

COS  I  smp 
=  tan  I  coip  +  1. 
COS  t  =  —  tan  Z  cot  j3. 
But       _p-90°-d 
.*.  cot  j3  =  tan  d, 
and      cos  i  =  —  tan  I  tan  d 


sin  2-  -4  =  ■v/sin(s— J)sin(s— c)csc  h  csc  c. 
Z  =  0. 
d=0. 

FZ  =90° -I  =90°  =  e. 
Pif  -  dO°-d  =  90°  =  6. 
^ir=  90° -?i  =  a. 
A  =  t. 
s-b  =  ^  (90°  -  A). 
s-c  =  ^  (90°  -  ^). 
csc  b  =  csc  90°  =  1. 
csc  c  =  csc  90°  =  1. 
Substitute 

sin  ^  i = Vsin  J  (90°-  h)  sin  J  (90°-;^) 
=  sin  J  (90° -/i). 
sin  ^  =  sin  (90°  -  h). 
t=90°-h 
=  z. 

(iii.) 
Zor(?=90°. 

PZ=90°-2  =  0°  =  a. 
.•.  no  triangle  will  be  made. 
.•.  answer  indeterminate. 

Pir=  90° -cZ=0°  =  6. 
/.  no  triangle  formed. 
.*.  result  indeterminate. 

23.  What  does  the  general  for- 
mula for  the  azimuth  of  a  celestial 
body,  in  §  68,  become  when  t  =  90° 
=  6  hours  ? 

When  t  =  90°,  m  =  0,  and  we  have 
a  right  spherical  triangle  with  the 
two  legs  given  to-  find  the  angle 
opposite  one  of  the  legs. 


TEACHERS     EDITION. 


161 


Then  by  substituting  the  values 
of  the  given  parts  in  the  formula, 

tan  B  =  tan  h  esc  a, 
we  have 

tan  a  =  tan  (90°-c?)  esc  (90°-Z) 
or     tana  =cot(Zsec^. 

tan  a      cot  a     sec  i 
and  cot  a  =  tan  d  cos  Z. 

24.   Show   that  the  formulas  of 
§  69,  if  t  =  90°,  lead  to  the  equation 
sin  Z  =  sin  A  esc  cZ ;  and  that  if  cZ  =  0°, 
they  lead  to  the  equation 
cos  I  =  sin  h  sec  t. 


I.  If<  =  90°. 

From  g  69, 

sin  ^  =  cos  n  cos  MQ. 

sin  d=  cosm  cos  MQ. 
Divide  (1)  by  (2), 


II.  If  (2=0°. 
From  §  69, 

cos  I  =  COS  m  sin  h  esc  c?.    (3) 
cos  t 


(1) 

(2) 


sin  ^ 


cos  n 


sin  cZ     cosm 

but  now    n  =  ZP=90°- 

and  m  =  0°. 

sin  h       •     7 

.'. =  sm  Z. 

sin  d 

.'.  sin  Z  =  sin  A  esc  cZ. 


tancZ  = 

tanm 

Multiply  (3)  by  (4), 

cos  Z  cos  t      cos  m  sin  h 


(4) 


tanm 

cos 

d 

cos  Z  cos  ^  = 

sin  m  sin 
cos  d 

_^ 

But  if    d  = 

0°, 

PM= 

90°, 

and           m  = 

0. 

.-,  cos  Z  cos  ^  = 

sin  h. 

.-.  cosZ  = 

sin  h 

sec 

<. 

25.  Given  latitude  of  place  52° 
30^  16^^  declination  of  star  38°,  its 
hour  angle  28°  17^  15^^  find  its 
altitude. 


Given       P^=90°-Z. 
PM=  90°  -  d. 
ZPM=  t ; 
required      ZM=  90°  -  h. 
Let  PQ  =  m. 

By  Napier's  Rules, 

tan  m  =  cot  d  cos  cZ. 


162 


TRIGONOMETRY. 


log  cot  fZ  =  0.10719 
log  cos  t  =  9.94477 
log  tan  m       =  10.05196 

m  =  48°  25^  10^^ 

sin  h  =  sin  (Z  +  m)  sin  d  sec  m, 

log  sin  (Z+m)=  9.99206 
log  sin  d  =  9.78934 
log  seem        =0.17804 

log  sin  h         =  9.95944  - 10 
h  =  65°  37^  20^. 

26.  Given  latitude  of  place  51° 
19''  20'''',  polar  distance  of  star 
67°  59^  5^^  its  hour  angle  15°  8^ 
12'''';  find  its  altitude  and  its  azi- 
muth. 


Given  PM=  polar  distance  of  star. 
ZPM=  hour  angle  of  star. 
PZ  =  co-latitude  of  observer. 
Find  PZM=  azimuth  of  star, 
and     DM  its  altitude. 
Let       d=90°-  PM. 
Let  PQ  =  m. 

tan  m  =  cot  d  cos  t. 
sin  h  =  sin  {I  +  m)  sin  d  sec  m. 
tan  a  =  sec  {I  +  m)  tan  t  sin  m. 
I  =  51°  19^  20^^. 


c?  =  90°  -  (67°  59^  5^0- 

=  22°  0^  55^^. 
t  =  15°  8^  12^^. 

log  cot  d        =  10.39326 

log  cos  t         =    9.98466 

^  log  tan  m       =10.37792 

m  =  67°  16^  22^^ 

log  sin  (l+m)  =  9.94351 
log  sin  d  =  9.57387 
log  sec  m        =  0.41302 

log  sin  h        =  9.93040 

h  =  58°  25^  15^^. 

log  sec  {l+m)  =  0,32001 
log  tan  t  =  9.43218 
log  sin  m       =  9.96490 

log  tana        =9.71709 
a  =  152°  28^. 

27.  Given  the  declination  of  a 
star  7°  54^  its  altitude  22°  45^  12^^ 
its  azimuth  129°  45^  37^^ ;  find  its 
hour  angle  and  the  latitude  of  the 
observer. 

sin  t  =  8ma  cos  h  sec  d. 


=  9.88577 
=  9.96482 
=  0.00414 


log  sin  a 

log  cos  h 

colog  cos  d 

log  sin  t         =  9.85473 
t  =  45°  42^ 

tan  m  =  cot  d  cos  t. 
cos  n  =  cos  m  sin  h  esc  d. 
I  =90°-(m±7i). 

log  cot  d        =  10.85773 
log  cos  t         =    9.84411 

log  tan  m       =  10.70184 

m  =  78°  45^  45^^. 


TEACHERS     EDITION. 


163 


log  cos  TO 

log  sin  h 
log  CSC  d 

log  cos  n 


=  9.28976 
=  9.58745 
=  0.86187 


=  9.73908 
n  =  56°  W  39^^. 

m-n  =  12°    V    Q'\ 

90°  _  (m  -  w)  =  67°  58^  54^^ 

.-.  Z  =  67°  58^  54^^. 

28.  Given  the  longitude  u  of  the 
sun,  and  the  obliquity  of  the  eclip- 
tic e  =  23°  27^ ;  find  the  declination 
d,  and  the  right  ascension  r. 


In  the  figure  let  P  represent  the 
pole  of  the  equinoctial  A  VB,  S  the 
position  of  the  sun,  and  Q  the  pole 
of  the  ecliptic  EVF. 
Then         VS  =  u. 
VB  =  r. 
8R  =  d. 
RVS=e. 

Then  in  the  right  triangle  B  VS, 
by  [38], 

sin  SB  =  sin  VS  X  sin  B  VS, 

sin  d 


or 


sm  u  sm  e. 


Also  by  [39], 

cos  BVS=  tan  B  Fcot  VS, 
or  cos  e  =  tan  r  cot  u. 

tan  r  =  tan  u  cos  e. 


29.  Given  the  obliquity  of  the 
ecliptic  e  =  23°  27^,  the  latitude  of 
a  star  51°,  its  longitude  315°;  find 
its  declination  and  its  right  ascen- 
sion. 

In  Fig.  47,  given 

VT=  315°  or  -  45°, 

TM=  51°, 

i2Fr=  23°  27^         , 

to  find         VB^r 

and  BM=  d. 

In  right  triangle  VTM, 

cos  VM=  cos  FT'cos  TM, 
and    tan  ili'FT^  tan  MT  esc  FT. 

log  cos  315°  =  9.84949 
log  cos  51°  =  9.79887 
log  cos  VM       =  9.64836 

VM=  63°  34^  36^^. 


log  tan  51° 
log  CSC  315° 

log  tan  MVT 
MVT- 


=  10.09163 
=    0.15051  (n) 

=  10.24214  in) 

-  (60°  12^  14.5^0- 


In  right  triangle  B  VM, 
BVM=BVT+  TVM 

=  23°  27^-(60°  12^  14.5^0 
=  -  (36°  45^  14.5^0- 
By  [38], 

sin  BM=  sin  FJf  sin  B  VM. 

log  sin  VM  =  9.95208 
log  sin  i?  FIT  =9.77698 
log  sin  BM       =  9.72906 

BM=d=2>2°2¥l2^'. 


164 


TRIGONOMETRY. 


Also,  by  [41], 

sin  VB  =  tan  BM  cot  E  VM. 

.   log  tan  i^lf      =9.80257 
logcoti^FJlf    =  0.12677  (n) 

log  sm  VB         =  9.92934  {n) 

Fi^= -(58°  11^  43^0- 
.-.  Fi2  =  360° -58°  IIMS^^ 

=  301°48M7^^ 

30.  Given  the  latitude  of  a  place 
44°  50^  14^^  the  azimuth  of  a  star 
138°  58^  43^^  and  its  hour  angle 
20°;  find  its  declination. 


Given     c  =  90°  -  44°  50'  14^' 

=  45°  9'  46^'. 

A  =  138°  58'  43''. 

B  =  20°. 

1{A-B)==  59°  29'  22". 

'l{A-\-B)  =  79°  29'  22". 

J  c  =22°  34'  53". 

log  cosJ(^-5)  =  9.70560 

colog  cos  jU+^)  =  0.73893 

log  tan  J  c  =9.61897 

log  tan  J  (a +  5)  =  0.06350 

Ha  +  &)  =  49°  10'  26". 


log  sin  J  {A-B)  =  9.93528 

colog  %m^{A+B)  =  0.00735 

log  tan  J  c  =9.61897 

log  tan  J(a- 6)  =  9.56160 

^(a-6)  =  20°  1'21.5". 

.-.  a  =  69°  11'  48". 

90°  -  69°  11'  48"=  20°  48'  12". 

31.  Given  latitude  of  place  51° 
31'  48",  altitude  of  sun  west  of  the 
meridian  35°  14'  27",  its  declina- 
tion +21°  27' ;  find  the  local  ap- 
parent time. 

Byi67, 

P^=90°-?, 
■     Pir=  90° -(Z  =  p, 
ZM=  90°  -  A ; 
required  t  =  ZBM. 

p  =  68°  33'. 

J(Z  +  A+^)  =  77°39'37.5". 
J(Z-/i+p)  =  42°  25' 10.5". 


log 

log 

colog 

colog 

log 


15 


cosJ(^+P  +  ^)=    9.32982 

sin  J(Z+_p-A)=    9.82901 

cosZ  =   0.20614 

sinp  =    0.03117 

2)19.39614 

sin  I «  =    9.69807 

J  ^  =  29°  55'  55.5". 

t  =  59°  51'  51". 

=  3  h.  59  min.  27f  sec.  p.m. 


32.  Given  latitude  of  place  ?,  the 
polar  distance  p  of  a  star,  and  its 
altitude  h  :  find  its  azimuth  a. 


TEACHERS     EDITION. 


165 


I 


Altitude  =  ZM=  90°-  h. 

Co- latitude        =  FZ  =  90°- 1. 
Polar  distance  =  FM 

=  90° -d=p. 
Azimuth  =  FZM  oi  a. 


cos  ^A=  Vsin  s  sin  (s—  a)  esc  b  esc  c. 


Let 


Then 
sin  s 


A^  FZM  or  a, 
a=p, 

b  =  90°-  h, 
c  =  90°-  I 


=  sm[90°-}  (l+h-p)] 
=  cos  J  (h  +  l—p). 

sin  (s-a)  =  sin  [90°-  H^+^+P)] 
=  cos  J(/i  +  ^+p). 

CSC  b        =  CSC  (90°—  A)  =  sec  h. 

esc  c  =  esc  (90°—  l)  =  sec  I. 
.'.  cos  I  a  = 

Vcos|(_p+A+^)cos|(/i+^-p)sec^secA 


SUEYETI.N-G. 


Exercise  I.     Page  143. 

1.   Required  the  area  of  a  triangular  field  whose  sides  are  respectively 
13,  14,  and  15  chains. 


Area  =  Vs{s  —  a)  (s  —  6)  (s  —  c). 

s  =  i(13  +  14  +  15)  =  21,  s- 5  =  21 -14  =  7,     ■ 

s-a  =  21 -13  =  8,  s-c  =  21 -15  =  6. 

Area  =  V21  x  8  x  7  X  6  =  VS^  x  7^  X  2*  =  3  X  7  X  2^ 

=  84  sq.  ch.  =  8.4  A.  =  8  A.  64  p. 

2.   Required  the  area  of  a  triangular  field  whose  sides  are  respectively 
20,  30,  and  40  chains. 


Area  =  V45  x  25  x  15  x  5  =  V3=*  X  5^  =  3  x  5 V3  x  5 

=  75  VTS  =  290.4737+. 
290.4737 sq.  ch.  =  29.04737  a.  =  29  a.  7.579  p.  =  29  a.  7f  p.,  nearly. 

3.  Required  the  area  of  a  triangular  field  whose  base  is  12.60  chains, 
and  altitude  6.40  chains. 

Area  =  J  base  X  altitude. 

Area  =  J  X  12.6  X  6.4  =  40.32  sq.  ch.  =  4.032  A.  =  4  A.  b^^j  p. 

4.  Required  the  area  of  a  triangular  field  which  has  two  sides  4.50 
and  3.70  chains,  respectively,  and  the  included  angle  60°. 

Area  =  ^bc  sin  A. 

Area  =  ^  x  4.5  x  3.7  X  0.866  =  7.20945  sq.  ch.  =  0.7209  a. 
=  115/oP.,  nearly. 

5.  Required  the  area  of  a  field  in  the  form  of  a  trapezium,  one  of 
whose  diagonals  is  9  chains,  and  the  two  perpendiculars  upon  this  diag- 
onal from  the  opposite  vertices  4.50  and  3.25  chains. 

Area  =  J  X  9  (4.5  +  3.25)  =  34.875  sq.  ch.  =  3.4875  a. 
.  =  3  A.  78  p. 


168  SURVEYING. 


6.  Required  the  area  of  the  field  ABCDEF (¥ig.  19),  if  ^^=9.25 
chains,  FF^=6A0  chains,  BE=  13.75  chains,  DD^=  7  chains,  DB=  10 
chains,  CC^=  4  chains,  and  AA^=4:.15  chains. 

2area^i^.£'     =    6.4x9.25         =    59.2 
2  area  BDFA  =  13.75  (4.75  +  7)  =  161.5625 
2zxQ2iBDC     =10x4  =40 


2  area  ABC  DBF  =  260.7625 

area  ABCDFF  =  130.38125 

130.38125  sq.  ch.  =  13.038125  a.  =  13  a.  6j\  p. 

7.  Required  the  area  of  the  field  ABCDEF  (Fig.  20),  if  AF^--^  4  chains, 
FF'=  6  chains,  EE'=  6.50  chains,  AE^=  9  chains,  AD  =  14  chains,  AQ' 
=  10  chains,  AB^=  6.50  chains,  BB^=  7  chains,  CC^=Q.7o  chains. 

2  area  Tli^i^''      =4x6  =24 

2areai^''^^^i^=5(6  +  6.5)       =    62.5 
2&YesLEE'B     =6.5x5  =    32.5 

2  area  ABB^     =  6.5  x  7  =    45.5 

2  area  BCC'B^  =  3.5  (7  +  6.75)  =    48.125 
2areaaDC^       =6.75x4  =    27 

2  area  ABCDEF  =  239.625 

area  ABCDEF  =  119.8125 

119.8125  sq.  ch.  =  11.98125  a.     =  11  a.  157  p. 

8.  Required  the  area  of  the  field  AGBCD  (Fig.  15),  if  the  diagonal 
AC=  5,  BB^  (the  perpendicular  from  B  to  AC)  =  1,  DD^  (the  perpen- 
dicular from  D  to  AC)^  1.60,  EE'=  0.25,  FF'=  0.25,  GG'=  0.60,  HM' 
=  0.52,  KK'=OM,  AE^  =  0.2,  E^F^^O.bO,  F^G^=0A5,  G^H^=0A5, 
S^K'^'-Om,  and  ^^^  =  0.40. 

2sirea,ADCB       =5(1+1.6)  =13. 

2  area  AEE'        =  0.25  x  0.2  =  0.05 

2  3.vesiEE'F'F    =0.5(0.25  +  0.25)=  0.25 

2&reiiFF'G'G    =0.45(0.25  +  0.6)=  0.3825 

2  area  (?G^i7^^  =  0.45(0.6 +0.52)  =  0.504 

2  area  ^.ff^ir'ir=  0.6  (0.52 +  0.54)=  0.636 

2  9^vesi,KK'B        =0.4x0.54  =  0.216 

2  area  ADCBKHGFE  =  15.0385 

Sivesi  ADCBKHGFE  =    7.51925. 


TEACHERS     EDITION. 


169 


9.  Required  the  area  of  the  field  AOBCD  (Fig.  16),  iiAD  =  Z,AC 
=  5,  AB=Q,  angle  i)J.a=  45°,  angle  ^4C=30°,  ^^^=0.75,  AF^=2.2b, 
AE=  2.53,  AG'=  3.15,  EE'=  0.60,  FF'=^  0.40,  and  0Q'=  0.75. 


2  area  ^1)05  =  3  X  5  X  0.7071  +  5  x  6  x  0.5 
2q.xq2.EGB    =0.75x3.47 


=  25.6065 
=    2.6025 


2s,Yea,AI)CBGE  =28.2090 

2  area  AEFH=  0.75  x  0.6  +  1.5(0.6  +  0.4)  +  0.4  x  0.28  =    2.062 


2  area  ADCBGHFE 
area  ADCBGHFE 


=  26.147 
=  13.0735. 


10.    Determine  the  area  of  the  field  ABCD  from  two  interior  stations 
F  and  P^  if  PP'=  1.50  chains, 

angle  P^PB^  3°  35^ 
P''P^  =  113°  45^, 
P^PD  =  165°  40^ 
P'PC  =  303°  15^ 


angle   PP'G  =    89°  35^ 

PP^P  =  185°  30^ 

PP^A  =  309°  15^ 

•  PP^i)  =  349°  45^ 


Area  =  A  PAD  +  A  PCD  +  A  PPC+  A  PAB. 


ZPP'D=  10°  15^ 
ZPDP'=  4°  5^ 
ZPP^P  =  174°  30^ 


Z  PPM  =  50°  45^ 
ZP^P^  =  15°  30^ 
ZPBF=    1°55^ 


PD  = 


PP'  sin  PP^P> 


sin  PDF' 

log  PP^  =  0.17609 

log  sin  PP^D  =  9.25028 

colog  sin  PDP^  =  1.14748 

log  PD  =  0.57385 


ZPP'C- 
ZPCP^-- 


PP^  sin  PPM 


89°  35^, 
33°  40^. 


PA  = 

sin  PAF^ 

log  PP^  =  0.17609 

log  sin  PPM  =9.88896 

colog  sin  PAP^=  0.57310 

log  PA  =  0.63815 


PC  = 


PP'smPP'C 


sin  PCP^ 

log  PP^  =  0.17609 

log  sin  PP^C  =  9.99999 

colog  sin  PCP^  =  0.25621 

log  PC  =  0.43229 


PB 


PP'  sm  PP'B 


sm  PBP^ 
log  PP^  =  0.17609 

log  sin  PP^B  =  8.98157 
colog  sin  PBP^  =  1.47566 

log  PB  =  0.63332 


Z  ^Pi)  =  51°  55^  ZZ>PC=  137°  35^  ZPPC=60°20^  Z^PP  =  110°10^ 


170 


SURVEYING. 


2  area  FAD  =  PDxPA  sin  APD. 
log  PD  =  0.57385 

log  PA  =  0.63815 

log  sin  APD  =  9.89604 

log  2  area       =  1.10804 
2  area  P^i)  =  12.825. 

2  area  P^P  =  PAxPB  sin  ^PP. 

log  P^  =  0.63815 

log  PB  =  0.63332 

log  sin  APB  =  9.97252 

log  2  area       =  1.24399 
2areaP^P  =17.538. 


2  area  PCZ)  =  PD  x  P(7sin  PPC. 
log  Pi)  =  0.57385 

log  PC  =  0.43229 

log  sin  DPC  =  9.82899 

log  2  area       =  0.83513 
2  Sires.  PCD    =6.8412. 

2  area  PBC=  PCx  PB  sin  PPC. 
log  PC  =  0.43229 

log  PB  =  0.63332 

log  sin  PBC  =  9.93898 

log  2  area       =  1.00459 
2areaPP(7   =10.106. 


2  A  P^P>  =  12.825 
2  A  PCD  =  6.841 
2  A  PBC  =  10.106 
2APAB  =17.538 


2ABCD    =47.310 
ABCD    =  23.655  sq.  ch. 

23.655  sq.  ch.  =  2.3655  a.  =  2  A.  58^-  p.,  nearly. 

11.   Determine  the  area  of  the  field  ABCD  from  two  exterior  stations 
P  and  P\  if  PP^=  1.50  chains. 

angle   P'PB  =    41°  10^  angle   PP'D  =    66°  45^ 

P'PA  =    55°  45^  PP'C  =    95°  40^ 

P^PC  =    77°  20^  PP^B  =  132°  15^ 

P^PD  =  104°  45^  PP^A  =  103°    0^ 


Area  =  (A  P^CP  +  A  P'CD)  -  (A  P^^P  +  A  P'AD). 


ZP''PP  =  41°  10^ 
ZPBP'=  6°35^ 
Z  P^P^  =  55°  45^ 


ZP^PP  =  104°  45^ 


Z  PDP'  = 
ZPAP'  = 


P'B  = 


PP'  sin  P^PP 


sin  PBP^ 

log  PP^  =  0.17609 

log  sin  P^PP=  9.81839 

colog  sin  PBP^  =  0.94063 


log  P'B 


0.93511 


8°  30^ 
21°  15^ 

P'D  = 


ZP^PC=  77°  20^ 
ZPCP'=   7°   0'. 


PP^  sin  P^PD 


sin  PPP^ 
log  PP^  =  0.17609 

log  sin  P^PD  =  9.98545 
colog  sin  PPP^  =  0.83030 

log  P'D         =  0.99184 


TEACHERS     EDITION. 


171 


P'G 


FF'  sin  P'PQ 


sm  PCP^ 
log  PP'  =  0.17609 

log  sin  P^PC=  9.98930 
colog  sin  PCP^  =  0.91411 

log  P^C  =  1.07950 


p^^  _  PP'  sin  P'PA 
sm  P^P-' 

log  PP^  =  0.17609 

log  P'PA       =  9.91729 

colog  P^P''       =  0.44077 

log  P'A         =  0.53415 


ZPP^C=36°35^ 
ZCP''i)  =  28°  55^ 


Z^P''P  =  29°15^ 
Z  ^P^i)  =  36°  15^ 


2  area  P^CP  =  P^Cx  P^B  sin  PP'^O. 
log  P^C  =  1.07950 

log  P'B         =  0.93511 
log  sin  PP^ (7  =9.77524 


log  2  area 
2  area  P^CP 


1.78985 
61.639. 


2  area  P'^CP  =  P^Cx  P^Psin  CP^P. 
log  P'C         =  1.07950 
log  P'B         =  0.99184 
log  sin  CP^n  =  9.68443 

log  2  area       =  1.75577 
2  area  P^CP  =  56.986. 


2  area  PMP  =P^P  x  P'A  sin  ^P^P. 

log  P^P  =0.93511 

log  PM  =  0.53415 

log  sin  ^P^P=  9.68897 


i 


log  2  area 
2  area  P^AB 

2  A  P^CP 
2  A  P^CP 

=  1.15823 
=  14.396. 

=    61.639 
=    56.986 

118.625 
34.248 

2  ABCD 
ABCD 

=    84.377 
=   42.1885 

2  area  P^AD  ^P^A  x  P^  D  sin  AP^D. 
log  PM  =  0.53415 

log  P^D         =  0.99184 
log  sin  ^P^P=  9.77181 

log  2  area       =  1.29780 
2  area  P^AD  =  19.852. 


2  A  P'AB 
2  A  P'' J.P 


=  14.396 

=  19.852 

34.248 


42.1885sq.ch.  =  4.21885  A. 

=  4  A.  35  p.,  nearly. 


172 


SURVEYING. 


Exercise  II.     Page  152. 


1. 


iV^. 

S. 

U. 

w. 

M.D. 

li.A. 

S.A. 

1 

2 
3 

4 

5 

S.  750  E. 
S.  150  E. 

S.  75^  W. 
N.  450  E. 
N.  450  W. 

6.00 
4.00 
6.93 
5.00 

5.19.^ 

3.54 
3.67 

1.55 

3.86 

1.80 
-1-^9- 

5.79 
-5-80- 

1.04 
3.54 

6.70 
-6-69- 

3.67 

5.79 
6.83 
0.13 
3.67 
0 

5.79 

12.62 
6.96 
3.80 
3.67 

13.4520 
13.4689 

8.9745 
48.7132 
12.5280 

21.647  sq.  ch.  =  2.1647  A.  =  2  A.  26  p.,  nearly. 

26.9209 

70.2157 
26.9209 

43.2948 
21.6474 

2. 


m 

S. 

E. 

w. 

M.D. 

N.A. 

S.A. 

1 

N.  450  E. 

10.00 

7.07 

.  .  . 

7.07 

.  .  . 

7.07 

7.07 

49.9849 

2 

S.  750  E. 

11.55 

.  .  . 

2.99 

11.16 

.  .  . 

18.23 

25.30 

75.6470 

3 

S.  15°  W. 

18.21 

.  .  . 

17.59 

.  .  . 

4.71 

13.52 

31.75 

558.4825 

4 

N.450  W. 

19.11 

13.51 

13.52 

0 

13.52 

182.6552 

..... 

232.6401 

634.1295 
232.6401 

200.74  Bq.ch.=S 

!0.074  A.  =  20  A.  12  P.,  nearly 

401.4894 

200.7447 

TEACHERS     EDITION. 


irs 


3. 


N. 

S. 

E. 

W. 

M.D. 

C5 

N.A. 

S.A. 

1 

^ 

N.150E. 

3.00 

2.90 

... 

0.78 

0.78 

0.78 

2.2620 

2 

N.  750  E. 

6.00 

1.55 

5.79 
-5-80- 

6.57 

7.35 

11.3925 

3 

s.  150  W. 

6.00 

5.80 

.  .  . 

1.55 

5.02 

11.59 

67.2220 

4 

N.  750  W. 

5.20 

1.35 

5.02 

0 

5.02 

6.7770 

20.4315 

67.2220 
20.4315 

23.395  sq.  ch.  =  5 

J.3395  A 

L.  -  2  A.  54  P., 

nearly. 

46.7905 

23.3953 

4. 


N. 

S. 

E. 

W. 

M.D. 

N.A. 

S.A. 

1 

2 

3 
4 
5 
6 

N.  890  45'  E. 
S.    70  00'  W. 

S.  28°  00'  E. 

S.   0045'E. 

N.  84°  45'  W. 
N.    2030'W. 

4.94 
2.30 
1.52 
2.57 
5.11 
5.79 

0.00 

-e-02- 

0.45 
-e-47- 

5.76 
-5-78- 

2.29 
-2-28- 

1.34 

2.58 
-2-57- 

4.93 
-4-94- 

•  .  . 

0.71 

0.02 
-0-03- 

0.29 
-0-28- 

5.10 
-5-09- 

0.27 
-0-25- 

4.93 
4.64 
5.35 

5.37 
0.27 
0 

4.93 
9.57 
9.99 
10.72 
5.64 
0.27 

2.5380 
1.5552 

21.9153 
13.3866 

27.6576 

2-94.^  A.  —  2  A.  I.'il  p..  nparlv. 

4.0932 

62.9595 
4.0932 

58.8663 
29.4332 

174 


SURVEYING. 


5. 


iV. 

S. 

E. 

W. 

N 

51045 

W. 

2.39 

1.48 

.  .  . 

.  .  . 

1.88 

S. 

85° 

w. 

6.47 

.  .  . 

0.56 

6.45 

S. 

550  10' 

w. 

1.62 

0.93 

1.33 

1.49 

.  .  . 

9.66 

1.48 

0.01 

N. 

8. 

E. 

w. 

M.D. 

N.A. 

S.A. 

2 
3 

4 
5 
6 
1 

S.                        W. 

N.    30  45'  E. 
S.  66045'E. 
N".  15°        E. 
S.  82°  45'  E. 
S.     2oi5'E. 

6.39 
1.70 
4.98 
6.03 
9.68 

6.36 
-6-38- 

4.80 
-4-81- 

0.03 

-0-ei- 

0.67 

0.77 
-e-T6- 
9.69 

0.43 
-e-42- 

1.56 
1.29 

5.98 

0.39 
-9-38- 

9.65 
-9-66- 

9.65 
9.22 
7.63 
6.37 
0.39 
0 

9.65 

18.87 

16.88 

14.03 

6.76 

0.39 

120.0132 
67.3440 

0.2895 

11.3096 

5.2052 
3.7791 

8  S.qQ   A    —  8  A     ^1-p      TiParW 

187.3572 
20.5834 

20.5834 

•»'j« 

166.7738 
83.3869 

TEACHERS     EDITION. 


175 


6. 


iV. 

S. 

E. 

W. 

S.  81O20' W. 
N.  76°  30'  W. 

4.28 
2.67 

0.62 

0.65 

.  .  . 

4.23 
2.60 

0.65 
0.62 

6.83 

0.03 

N. 

S. 

JE. 

W. 

S.     7°        E. 
S.  27°        E. 
S.  103  30'E. 
N.  76^45' W. 

1.79 
1.94 
5.35 
1.70 

0.39 

1.78 
1.73 
5.26 

0.22 
0.88 
0.98 

1.65 

8.77 
0.39 

2.08 
1.65 

.  .  . 

8.38 

0.43 

1 
2 
3 
4 

N'. 

S. 

E. 

w. 

M.D. 

N.A. 

S.A. 

S.               W. 

N.    5°        E. 
S.  87°  30'  E. 
S.               E. 

8.68 
5.54 

8.65 

0.03 

0.24 
8.38 

0.79 
-e-76- 

5.55 
-6-53- 

0.46 
-0-43- 

6.80 
-6-83- 

6.80 
6.01 
0.46 
0 

6.80 

12.81 

6.47 

0.46 

110.8065 

0.2040 

1.5528 

3.8548 

5.2597  A.  =  5  A.  42  P..  nearlv. 

110.8065 
5.6116 

5.6116 

105.1949 

52.597 

176 


SUEVEYING. 


7. 


2f. 

S. 

E. 

W. 

M.n. 

mA. 

S.A. 

3 

4 
1 
2 

S.     5O00'E. 
N.  88^30' E. 
N.    6°  15'  W. 
S.  81O50' W. 

5.86 
4.12 
6.31 
4.06 

0.12 
-0-11- 

6.28 
-6-27- 

5.83 
6-84- 

0.57 
-e-58- 

0.53 
-6-51- 

4.14 
-4-12- 

0.67 
-9.69- 

4.00 
-4-02- 

0.53 

4.67 
4.00 
0 

0.53 

5.20 
8.67 
4.00 

0.6240 
54.4476 

3.0899 
2.2800 

55.0716 
5.3699 

5.3699 

5 

U85  a.=  2a.  78 

P.,  ne< 

irly. 

49.7017 
24.8508 

8. 


■ 

iV^. 

S. 

E. 

r. 

M.D. 

N.A. 

S.A. 

3 
4 
1 
2 

S.    3°  00'  E. 
E. 
N.   5030'W. 
S.  82°  30'  W. 

5.33 
6.72 
6.08 
6.51 

0.03 

-e-ee- 
6.0s 

-6-65- 

5.29 
-5-32- 

0.82 
-6-85- 

0.28 

6.73 
-6-72- 

0.57 
-e-58- 
6.44 
-6-45- 

0.28 
7.01 
6.44 
0 

0.28 

7.29 

13.45 

6.44 

0.2187 
81.7760 

1.4812 

5.2808 

81.9947 
6.7620 

6.7620 

J.761  A.  =  3  A.  12 

2  P.,  m 

iarly. 

75.2327 
37.6163 

TEACHEES     EDITION. 


177 


9. 


1 

2 
3 

4 

5 

1 

iV. 

S. 

E. 

W. 

M.n. 

c5 

N.A. 

S.A. 

N.20°00'E. 
N.  73°  00'  E. 
S.  450  15'E. 
S.  38°  30'  W. 
"Wanting. 

4.62i 
4.16i 
6.18i 
8.00 

4.35 
1.22 

5.04 

4.35 

6.26 

1.58 
3.98 
4.39 

4.98 
4.97 

1.58 
5.56 
9.95 
4.97 
0 

1.58 

7.14 

15.51 

14.92 

4.97 

6.8730 
8.7108 

25.0488 

67.4685 
93.3992 

1         ■  ■ 

40.6326 

160.8677 
40.6326 

( 

5.012  A.  =  6  A.  2  ] 

P.,  neai 

ly. 

120.2351 
60.1175 

10. 


N. 

S. 

E. 

W. 

M.D. 

c5 

N.A. 

S.A. 

6 

7 
8 
9 
1 
2 
3 
4 
5 

N.  32°  00'  E. 

s.  75050'E. 

S.  140  45' "W. 
S.  79"15'E. 

s.  3000'E. 

S.  860  45'"W. 
S.  37°00' W. 
N.  810  00'W. 
N.  610  00'W. 

8.68 
6.38 
0.98 
4.52 
4.23 
4.78 
2.00 
7.45 
2.17 

7.3 
-7-3 

1.1^ 

-1-1- 
1.0 

-1-0^ 

3 
3- 

1 
?- 
i 
3- 

1.58 
-1-56- 

0.95 

0.86 
-6-84- 
4.23 

-4-^- 
0.29 
-0-27- 

1.60 

4.61 
-4-69- 

6.20 
-6-19- 

4.44 
0.22 

0.25 

4.77 

1.20 

7.35 
-7-36- 

1.90 

4.61 
10.81 
10.56 
15.00 
15.22 
10.45 

9.25 

1.90 

0 

4.61 
15.42 
21.37 
25.56 
30.22 
25.67 
19.70 
11.15 

1.90 

33.7913 

12.7110 
1.9760 

24.3636 
20.3015 
21.9816 
127.8306 
7.4443 
31.5200 

48.4783 

233.4416 
48.4783 

{ 

).248  A.  =  9  A.  40 

P.,  nea 

rly. 

184.9633 

92.48 

178 


SUHVEYING. 


Exercise  HI.     Page  163. 


1. 


N. 

S. 

E. 

w. 

M.  D. 

N.A. 

S.A. 

BC 

S.  60°  E. 

4.000 

4.000 

4.000 

0 
3.464 

0 
3.464 

.... 

6.928 

2.000 

3.464 

•  .  > 

CD 

S.  30°  E. 

6.928 

6.000 

3.464 

.  .  . 

6.928 

10.392 

.... 

62.352 

DA 

N.  60O  W. 

8.000 

4.000 

.... 

6.928 

0 

6.928 

27.112, 

.... 

27.712 

69.280 
27.712 

20.784  sq.  ch.=  2.0 

784  A.  =  2  A.  12|  p.,  nearly. 

41.568 

20.784 

2. 


N. 

S. 

E. 

W. 

M.D. 

N.  A. 

S.A. 

AB 
BC 

N. 

N.  80°20'E. 

79.86 
121.13 

79.860 
20.338 

0 
119.410 

0 

119.410 

2428.56 

.... 

.  .  . 

119.410 

.  .  . 

CD 

S.  40°  00'  E. 

90.00 

.  .  . 

68.943 

57.851 

177.261 

296.671 

.... 

20453.39 

DM 

s.  55052'w. 

100-65 

.  .  . 

59.350 

.... 

81.289 

95.972 

273.233 

2696.33 

16216.38 

EA 

N.  730  41'  W. 

109.00 

28i)95 

.... 

95.972 

0 

95.972 



.... 

5124.89 

36669.77 
5124.89 

15772.44  jp.  =  9^  A.  92  P.,  nearly. 

31544.88 

15772.44 

teachers'  edition.  179 


Exercise  IV.     Page  161. 

1.  From  the  square  ABCB,  containing  6  a.  1  r.  24  p.,  part  off  3  A. 
by  a  line  EF  parallel  to  AB. 

6  A.  1  R.  24  E.  =  64  sq.  ch. ;   Voi  =  8  ch.  =  AB. 
3  a.  =30  sq.  ch. 

AB         8 

2.  From  the  rectangle  ABCD,  containing  8  a.  1  n.  24  P.,  part  off  2  A. 
1  E.  32  p.  by  a  line  ^i^  parallel  to  AI)=7  ch.  Then,  from  the  remain- 
der of  the  rectangle  part  off  2  a.  3  r.  25  p.  by  a  line  GE  parallel  to  IJB. 

8  A.  1  R.  24  p.  =  84  sq.  ch.  =  ABCB. 
2  A.  1  R.  32  p.  =  24.5  sq.  ch.  =  AEFD. 
2  A.  3  R.  25  p.  =  29.0625  sq.  ch.  =  EBHQ. 

AE  =  ^^^='-^^3.5ch. 
AD         7 

A^CD^84^^2ch. 

AD         7 

EB  =AB-AE  =12-  3.5  =  8.5  ch. 

^  EBHG  ^2jm25  ^  3  ^^  ^h..  nearly. 
EB  8.5 

3.  Part  off  6  A.  3  e.  12  p.  from  a  rectangle  ABCD,  containing  15  A. 
by  a  line  EF  parallel  to  AB ;  AD  being  10  ch. 

6  A.  3  R.  12  p.  =    68.25  sq.  ch.  =  ABFE. 
15  A.  =  150  sq.  ch.  =  ABCD. 

AB=^^^  =  '-^  =  15ch. 
AD         10 

^^  =  ^^^=68^  =  4.55ch. 
AB  15 

4.  From  a  square  ABCD,  whose  side  is  9  ch.,  part  off  a  triangle 

which  shall  contain  2  a.  1  r.  36  p.,  by  a  line  BE  drawn  from  B  to  the 

side  AD. 

2  A.  1  R.  36  p.  =  24.75  sq.  ch. 

AB  9 


180  SURVEYING. 


5.  From  ABCD,  representing  a  rectangle,  whose  length  is  12.65  ch., 
and  breadth  7.58  ch.,  part  off  a  trapezoid  which  shall  contain  7  A.  3  e. 
24  p.,  by  a  line  BE  drawn  from  B  to  the  side  DO. 

7  A.  3  E.  24  p.  =  79  sq.  ch. 

ABCD  =  12.65  X  7.58  =  95.887  sq.  ch. 

A  BCE=  95.887  -  79  =  16.887  sq.  ch. 

^55.     2BCE     2x16.887      .  .r-    .  , 

CE  = = =  4.456  ch.,  nearly. 

BC  7.58  ^ 

6.  In  the  triangle  ABC,  AB  =  12  ch.,  AC=  10  ch.,  and  5C=  8  ch. ; 
part  off  1  A.  2  E.  16  p.,  by  the  line  DjE" parallel  to  AB. 

1  A.  2  E.  16  p.  ==  16  sq.  ch. 

CAB  =  Vl5x  3x5x7  =  39.6863  sq.  ch. 

CDE  =  CAB  -  ABED   =  39.6863  -  16  =  23.6863  sq.  ch. 

CAB  :    CDE  ::  CA^  :  CD' 

:  :  CB^  :  CE\ 

39.6863  :  23.6863  :  :  10^  :  CD\    .'.  CD  =  7.725  ch. 
:  :    82;  CE\    .-.  C^-6.18    ch. 
AD  =  CA-  CD  =10-  7.725  =  2.275  ch. 
BE  =  CB  -  CE  =    8  -  6.18    =  1.82    ch. 

7.  In  the  triangle  ABC,  AB=2Q  ch.,  AC  =  20  ch.,  and  BC=  16  ch. ; 
part  off  6  A.  1  E.  24  p.,  by  the  line  DE  parallel  to  AB. 

6  A.  1  E.  24  p.  =  64  sq.  ch. 

CAB  =  V31x5x  11x15  =  159.9218  sq.  ch. 

CDE^  CAB  -  ABED  =  159.9218  -  64  =  95.9218  sq.  ch. 

CAB  :    CDE  :  :  CA"  :  CD" 


:  :  (7^' 

:  CE 

2 

159.9218  ; 

;  95.9218  :  : 

202; 

cd\ 

162  : 

ce\ 

.  CD  =  15.49  ch. 

.  CE  =  12.39  ch. 
AD=  CA-  CD  =  20-  15.49  =  4.51  ch.,  nearly. 
BE  =  CB  -  CE  =  IQ  -  12.39  =  3.61  ch.,  nearly. 

8.  It  is  required  to  divide  the  triangular  field  ABC  among  three  per- 
sons whose  claims  are  as  the  numbers  2,  3,  and  5,  so  that  they  may  all 
have  the  use  of  a  watering-place  at  C;  AB  =  10  ch.,  AC=  6.85  ch.,  and 
CB  =  6.10  ch. 


teachers'  edition.  181 

Since  the  tr' angles  have  the  same  altitude,  they  are  to  each  other  as 
their  bases.  Hence  it  is  only  necessary  to  divide  the  base  10  into  the 
three  parts,  2  ch.,  3  ch.,  5  ch. 

9.  Divide  the  five-sided  field  ABCHE  among  three  persons,  X,  Y, 
and  Z,  in  proportion  to  their  claims,  X  paying  ^500,  Y  paying!  750, 
and  Z  paying  %  1000,  so  that  each  may  have  the  use  of  an  interior  pond, 
at  P,  the  quality  of  the  land  being  equal  throughout.  Given  AB  =  8.64 
ch.,  ^(7=8.27  ch.,  CZr=  8.06  ch.,  HU  =  6.S2  ch.,  and  ^^  =  9.90  ch. 
The  perpendicular  FD  upon  ^^-5.60  ch.,  FB^  upon  ^C=6.08  ch., 
FB^^  upon  CH^  4.80  ch.,  FB''^  upon  HE^bA^  ch.,  and  FB^^^^  upon 
EA  =  5.40  ch.  Assume  FH  as  the  divisional  fence  between  X's  and  Z's 
shares  ;  it  is  required  to  determine  the  position  of  the  fences  FM  and 
PiV  between  X's  and  Y's  shares  and  Y's  and  Z's  shares,  respectively. 

If  Pbe  joined  to  the  vertices,  the  field  is  divided  into  triangles,  whose 
bases  are  the  sides,  and  the  altitudes  the  given  perpendiculars  upon  the 
sides  from  P. 

AFB  =  8.64  X  2.80  =    24.1920  sq.  ch. 

BFC  =  8.27  X  3.04  =    25.1408 

CFH  =  8.06  X  2.40  =    19.3440 

JTFE  =  6.82  X  2.72  =   18.5504 

EFA  =  9.90  X  2.70  =   26.7300 

ABCHE  =113.9572 

The  whole  area  113.9572  sq.  ch.  must  be  divided  as  the  numbers  500, 
750,  1000,  or  as  2,  3,  4.     2  +  3  +  4  =  9. 

9  :  113.9572  :  :  2  :  25.3238  sq.  ch.  =  X's  share. 
:  :  3  :  37.9857  sq.  ch.  =  Y's  share. 
:  :  4  :  50.6476  sq.  ch.  =  Z's  share. 

FH'is  assumed  as  the  line  between  X's  and  Z's  shares.  Since  the  tri- 
angle FHE  is  less  than  X's  share  by  25.3238  - 18.5504  =  6.7734  sq.  ch., 
this  difference  must  be  taken  from  the  triangle  FEA.  The  area  of  FEM 
is  then  6.7734  sq.  ch.,  and  the  altitude  Pi)^^/^  =  5.40. 

,.  ^J/=2P^^=2xa773_4^  2.5087  ch. 
FB''^^  5.40 

PMA  =  FEA  -  FEM=  26.7300  -  6.7734  =  19.9566.  sq.  ch. 

Since  Y's  share  is  greater  than  FMA  (19.9566)  and  less  than  FMA 
+  FAB  (44.1486),  the  point  iV^is  on  AB. 


182  SURVEYING. 


Y's  share  diminished  by  PMA  equals  PAN;  that  is, 

FAN=  37.9857  -  19.9566  =  18.0291  sq.  ch. 

^^^2P4i^^2xiM291  =  6.439  ch. 
FD  5.60 

10.  Divide  the  triangular  field  ABO,  whose  sides  AB,  AC,  and  BC 
are  15,  12,  and  10  ch.,  respectively,  into  three  equal  parts,  by  fences 
EG  and  Di^  parallel  to  BC. 


ABC  =  \/18.5  X  3.5  X  6.5  X  8.5  ==  5U81169  sq.  ch. 
ADF  =  ^  of  59.81169  =  19.9372  sq.  ch. 
AEG  =  I  of  59.81169  =  39.8744  sq.  ch. 
ABC  :  AEG  ::  AB''  :  AE'' 

:  :  AC''  :  Zg'. 
59.81169  :  39.8744  :  :  15^ :  AE^.    :.  AE=  12.247  ch. 
:  :  122;  A^.     .',AG=    9.798  ch. 
ABC  :  ADF:  :  Tb"  :  AD" 

:  :  ZC'  :  AF\ 
59.81169  :  19.9372  :  :  15^  :  A&.     .-.  AD  =  8.659  ch. 
:  :  122  .  Jp\     ,-,AF  =  6.928  ch. 

11.  Divide  the  triangular  field  ABC,  whose  sides  AB,  BC,  and  AC 
are  22,  17,  and  15  ch.,  respectively,  among  three  persons,  A,  B,  and  C, 
by  fences  parallel  to  the  base  AB,.  so  that  A  may  have  3  a.,  B  4  a.,  and 
C  the  remainder. 


CAB  =  V27  X  5  X  10  X  12  =  127.2792  sq.  ch. 
CDG  =  CAB  -  ABGD  =  127.2792  -  30  =  97.2792  sq.  ch. 
CEF  =  CAB  -  ABFE  =  127.2792  -  70  =  57.2792  sq.  ch. 
CAB  :  CDG  :  :  M^  :  CG^ 
:  :  OT  :  C&. 

127.2792  :  97.2792  :  :  17^ :  C^.    :.  CG  =  14.862  ch. 

^2 


:  :  152 .  cj)\    ...  CD  =  13.113  ch. 
CAB:CEF::CB':CF^ 
:  :  GZ'  :  GF. 
127.2792  :  57.2792  :  :  I72 :  CF^.    .-.  CF=  11.404  ch. 

:  :  152 :  Oil    .■.CE=  10.062  ch. 


^- 


TEACHERS     EDITION. 


183 


Exercise  V. 

1.  Find  the  difference  of  level  of  two  places  from  the  following  field 
notes :  back-sights,  5.2,  6.8,  and  4.0 ;  fore-sights,  8.1,  9.5,  and  7.9. 

8.1  +  9.5  +  7.9  =  25.5 

5.2  4-6.8+4    =16 

9.5 

2.  Write  the  proper  numbers  in  the  third  and  fifth  columns  of  the 
following  table  of  field  notes,  and  make  a  profile  of  the  section. 


Station. 

+S. 

II.  I. 

-S. 

H.S. 

Remarks. 

B 

0 

1 

2 

3 
t.p. 

4 

5 

6 

6.944 

20. 

19.5 

21.3 

23.0 

22.3 

21.431 

20.4 

21.8 

24.1 

Bench  on  post  22 
feet  north  of  0. 

26.944 

7.4 

5.6 

3.9 

4.6 

5.513 

4.9 

3.5 

1.2 

3.855 

25.286 

Surface. 


3.  Stake  0  of  the  following  notes  stands  at  the  lowest  point  of  a  pond 
to  be  drained  into  a  creek  ;  stake  10  stands  at  the  edge  of  the  bank,  and 
10.25  at  the  bottom  of  the  creek.  Make  a  profile,  draw  the  grade  line 
through  0  and  10.25,  and  fill  out  the  columns  H.O.  and  C,  the  former  to 
show  the  height  of  grade  line  above  the  datum,  and  the  latter,  the  depths 
of  cut  at  the  several  stakes  necessary  to  construct  the  drain. 


184 


SURVEYING. 


Station. 


B 

0 
1 

2 

3 

4 

5 

6 

7 

8 

9 
10 
10.25 


+  S. 


6.000 


572 


H.I. 


-S. 


10.2 

5.3 

4.6 

4.0 

6.8 

7.090 

3.9 

2.0 

4.9 

4.3 

4.5 
11.8 


H.S. 


25 


H.G. 


20,8 
20.4 
20.0 
19.6 
19.2 
18.8 
18.4 
18.0 
17.6 
17.2 
16.8 
16.7 


0.0 
5.3 
6.4 
7.4 
5.0 
5.1 
6.2 
8.5 
6.0 
7.0 
7.2 
0.0 


Remarks. 


Bench  on  rock 
30  feet  west  of 
stake  1. 


10  10.25 


Date  Due 

^ 

DATE  DUE 

.  1 

201-5503 

PRINTED  IN  U.S.A. 

BOSTON  COLLEGE 


3  9031  01550242  0 


